Bí quyết ôn luyện thi đại học đạt điểm tối đa vật lý tập 1 lê văn vinh part 2
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V
(phdn tich va huang dan gidi
2nv
271.50
= 2cm
Buac song:'k= — =
507t
f
O
Vi M va O n a m trenCd u o n g trung true ciia hai nguon nen
P h u o n g trinh song tai mpt M la: u ^ = 2acos 507it-
2Kd^
P h u o n g trinh song tai mot O la:
27i.OA~
U Q = 2a cos 507tt - = 2acos(507rt-87:)
M va O d a o dong nguoc pha khi:
= > A ( P M / O = 8 7 i - ^ = (2k + l ) 7 t
^ d = 3,5>.-k)L = 7 - 2 k > A O = 8 = ^ k < - 0 , 5
Vgy: d„i„ « k ^ 3 , = - 1 =^ d^in = 9 ^ OM^i„ = ^ d ^ i ^ - O A ^ = 7T7cm
C h p n d a p an A
C a u 5 : Hai nguon ke't h o p Si,S2 each nhau mpt khoang 50mm tren m a t nuoc
phat ra hai song ke't h o p c6 p h u o n g trinhuj = U 2 = 2cos2007it(mm). Van
toe truyen song tren mat nuoc la 0,8 m/s. Diem gan nhat d a o dpng ciing pha
voi nguon tren d u o n g trung true eua S1S2 each nguon Si bao nhieu:
D. 24mm
A. 16nim
B. 32mm
C. 8 m m
^hdn tich v>d huang dan gidi
Giai 1:
P h u o n g trinh song tong quat tong h p p tai M la:
U M = 2a cos
r
d2-di
cos 2007lt - 7t
Voi M each deu Si, S2 nen di = d2.
Khi d o d2 - di = 0
=> A = 2a
cos
De M d a o dpng cung pha voi Si, S2 thi:
dij^
d i + d ^ ^ 2 k = ^ d i = d 2 = d = kX
Goi X la khoang each t u M den Si va S2:
d , = d , = d = Jx'2.
+
rsiS2 ^
= > 0 , 6 4 k ' ' - 6 , 2 5 > 0 = > k > 3,125 =>
= 4 = > d = 4A = 32 mm.
ChQH d a p an B.
Giai n h a n h bai nay n h u sau:
Diem M d a o dong ciing pha voi nguon Si nen phai each Si m p t doan d = kX
Nhin tu hinh ve ta tha'y:
d = k^>S,0^8k>25:^k>3,125:^k„i„ = 4 => d^,„ = 4X = 4.8 = 3 2 m m
Chgn B
;au 6 : Tren mat mot cha't long, c6 hai nguon song ke't h p p Oi, O2 each nhau
/ = 24em, d a o dong theo ciing mpt p h u o n g voi p h u o n g trinh
"oi = " o 2 = Aeoswt (t tinh bang s A tinh bang mm). Khoang each ngSn nhat
tu trung diem O eua O1O2 den cac diem n a m tren d u o n g trung true eua
O1O2 d a o dong ciing pha voi O b3ng q = 9cm. So diem d a o dong voi bien dp
bang O tren doan O1O2 la:
A. 18
B. 16
C. 20
D. 14
^hdn tich m huang dan gidi
P h u o n g trinh d a o dong tai mpt diem khi c6 giao thoa:
u = 2 A cos
d, - d ,
cos (Ot - 71
P h u o n g trinh d a o dong tai O:
u = 2Acos cot-- 27ta (voi / = 2a)
1°
J,"
P h u o n g trinh d a o dong tai M:
.
u = 2Acos cot 27rd^
Dp l^ch pha ciia M so voi O: Acp = — ( d - a )
^'
X
2?^
M d a o dong ciing pha vai O nen: Acp = — ( d - a) = 2k7i => d - a = k>.
X
Diem M gan O nhat thi: k = 1
=>>. = d - a = 7a^+q^ - a = 712^+9^
So cue dai tren O1O2:
-12
= 3cm
- | < k < i = > - 8 < k < 8 : k = ±8;+7;±6;±5;±4;±3;+2;±l;0
Co p17 cuepdai tren OiCh (ke ea O), vay c616 diem dao dpng voi bien d p hang O
C h n d a an B
C a u 7: H a i n g u o n song ket hop,
dat tai A va B each nhau 20cm dao d g n g theo
di2 = (8 + x)2 + 82; d22 = (8 - xf- + 8^
p h u o n g t r i n h u = acos(cjt) tren mat nuoc, coi bien do k h o n g d o i , buoc song
=> di2 - d22 = 32x => d i + d2 = 16x
X = 3cm. Gpi O la t r u n g d i e m ciia A B . M o t d i e m n a m tren d u o n g t r u n g true
T u (*) va (**) =>di = 8 x + l
A B , dao d p n g c i m g pha v o i cac n g u o n A va B, each A hoac B m p t do^n nho
di2 = (8 + x)2 + 82 = (8x + 1)2 => 63x2 = 128 :
nhat la
B.lOem
C.13.5cm
D.lScm
= 1,42 cm.
!U,
Cach 3
gidi
Ban CO the giai theo p h u o n g t r i n h hypeebol n h u sau:
Bieu thiic song tai A , B la: u = aeoscot
Trong do N la d i n h hypeebol
A M = B M = d (cm) > 10 cm
Bieu thiie song tai M :
U ; ^ = 2acos c o t -
k^in = 4
27id
ket hop v o i hai n g u o n eung pha nen
ON = a = - = - = lem
4
4
c
x
y
?
0
/' M
/
/
(
X'
a
Theo bai ra thi khoang each giua xx' v o i A B la y = 8cm
C a u 8 : T r o n g thf n g h i ^ m giao thoa tren mat chat long v o i 2 n g u o n A , B phat
song ket h o p nguoe pha nhau. Khoang each g i i i a 2 n g u o n la A B = 16cm. H a i
song t r u y e n d i c6 buoc song la 4em. Tren d u o n g thang xx' song song v o i
A B , each A B m o t doan 8cm, goi C la giao d i e m cua xx' v o i d u o n g t r u n g true
eiia A B . K h o a n g each ngan nhat t u C den d i e m dao d p n g v o i bien d p cue
dai n a m tren xx' la
B. l,50em.
^hdn
C. 2,15em.
tich v>d hucmg dan
= 2 o x = l,42cm
K h i : d i - d2 = (k + 0,5) X
D i e m M gan C nhat k h i k = 0
(*)
B. 30cm
^han
C. 40em
tich vd hu&ng ddn
de doan A M eo gia trj Ion nhat t h i
C
X
M
x'
M phai nam tren van eye dai bae 1
n h u h i n h ve va thoa m a n :
d 2 - d , = k ? t = 1.20 = 20(cm) (1).
di
D i e m M dao d p n g v o i bien d p eye tieu
d p n g eung pha. Bie't song do m o i nguon phat ra eo tan so'f = lO(Hz), v a n toe
Do M la m o t eye dai giao thoa nen
Cach 2
= IH
C a u 9 : Tren be mat chat l o n g c6 hai nguon ket hop A B each nhau 40cm dao
T a c o ;. = ^ = ^
= 20(em).
f
10
^
'
k + - X o j 8 2 + ( 8 + x)^-^8^+{8-xf
2)
M a t khae, do tarn giae A M B la tam
A
I
H
= l . J l + = l,42em
63
C h p n dap an A
A . 20em
tren xx' t h i M thupe cue tieu t h u nhat k = 0
M = d2; X = C M
• = l = > x = a, l + ^
a
tai A dao d o n g v o i bien dp cue dai. Doan A M c6 gia t r i Ion nhat la:
gidi
K h o a n g each ngan nhat t u C den d i e m dao d p n g v a i bien dp eye tieu nSm
Xet d i e m M :
Vi the'tu phuong trinh:
truyen song 2(m/s). Gpi M la m o t d i e m n a m tren d u o n g v u o n g goc v o i A B
D . 2,25cm
Cach 1 : G o i M la d i e m thoa man yeu cau va dat C M = x,
di - d2 = 0,5>. = 2 (cm)
X
T a c o : b^ =e^ - a ^ - 8 ^ - 1 ^ =63
C h p n dap an A
= di; B
,'' k = 0
v o i e la tieu
diem va e = OB = O A = — = — = 8em
2
2
k>3,33
d^i„ =4X = 4.3 = 12cm
A . l,42em.
voi
d u o n g cue tieu gan t r u n g true nha't
D i e m M dao d p n g c i i n g pha v o i n g u o n k h i
= 2 k T i = > d = kX = 3 k > 1 0 ^
=1
a
Xet d i e m M tren t r u n g true ciia A B :
A M
X
Chpn D
A.12cm
—
(**)
B
giae v u o n g tai A nen ta c6:
A M = d2 = V ( A B 2 ) + ( A M 2 ) =740^+d,2 (2)
D.SOem
gidi
(2) v a o (1) t a d u o c :
Thay
H a i n g u o n d a o d o n g k h a c b i e n d o v i t h e t a ti'nh b i e n d o t h e o c o n g t h u c :
+ d ,^ - d ] = 2 0 : • d i = 3 0 ( c m )
SJAO^
Cau
10:
T r e n b e m a t c h a t l o n g c 6 h a i n g u o n k e t hgrp A B each n h a u 1 0 0 c m
d a o d o n g c u n g p h a . B i e t s o n g d o m o i n g u o n p h a t r a c 6 t a n so f = 10(H2),
v a n toe t r u y e n s o n g 2(m/s). G o i M la m o t d i e m
B. 2 2 , 5 c m
C. 12cm
D. 30cm
In
(0)n
= 5^ + 3 ^ + 2 . 5 . 3 . C O S
(d2 - d , ) - ( ( p 2 -cpi)
4;
U
= 34 + I5V2
A = V34 + I5V2 c m
Chpn dap an B
^hdn tich vd hu&ng dan gidi
Taco
= 5^ + 3^ + 2 . 5 . 3 . C O S
nam t r e n d u o n g v u o n g g o c
v o i A B t a i A d a o d o n g v o i b i e n d p c u e d a i . D o a n A M c 6 g i a t r j n h o n h a t la:
A . 5,28cm
2n
A ^ = A t + A2 + 2 A , A 2 C o s
C h p n dap an B
Cau 12:
T r e n m a t m o t c h a t l o n g c 6 h a i n g u o n s o n g ke't h o p c i i n g p h a c 6
bien d o 7 m m va 8 m m dao d o n g v u o n g goc v o i mat t h o a n g chat long. N e u
;i = ^ = ^
= 20(cm).
f
10
cho rSng song t r u y e n d i v o i bien d o k h o n g thay d o i t h i tai m o t d i e m M
So v a n c u e d a i t r e n d o a n A B t h o a m a n d i e u k i e n :
each h a i n g u o n n h u n g k h o a n g d i = 12,75>i v a d 2 = 7,25X se c 6 b i e n d p d a o
- A B < d 2 - d , = kA.< A B .
d o n g la b a o n h i e u ?
Hay :
-AB
X
,
AB
-100 ,
100 _
< k<
o
20
K
20
_ ,
-5 < k < 5 ,
A . 7,5 m m
B. 1 m m
C . 15 m m
D . 56 m m
^hdn tick v>d hucmg ddn gidi
Suy ra : k = 0 , ± 1 , ± 2 , ± 3 , ± 4 .
Do h a i n g u o n eo b i e n d o k h a c n h a u n e n ta t i n h b i e n d o s o n g t o n g h p p t a i M
Vay
theo cong t h u e t o n g h o p dao d o n g d i e u hoa:
de doan
A M c6 gia t r i be
n h a t th'i M p h a i n a m t r e n d u o n g
A ^ = A f + A ^ +2A,A2Cos
eye d a i bac 4 ( k = 4 ) n h u h i n h v e
va thoa m a n :
= 7^ +8^
d2 - d j = k ? t = 4 . 2 0 - 8 0 ( c m )
+2.7.8.COS
2n
2n,
^(d2-d,)-({p2-(pi)
{U,75X-7,25X)-0
A.
M a t khac, d o t a m giac A M B la
= 7^ + 8^ + 2.7.8.cosn7: = 1
t a r n g i a c v u o n g t a i A n e n t a eo:
=> A = l c m
M B - d 2 -V(AB2) + (AM2)=7lOO^+di2(2)
Chpn dap an B
T h a y (2) v a o (1) ta d u o c :
7lOO^ + d i ^ - d , = 8 0 ^ d i
=22,5(cm)
Chpn dap an B
Cau
11:
T a i h a i d i e m A , B tren m a t chat l o n g c6 h a i n g u o n phat
song:
Uys^ = 5 c o s ( ( o t ) c m ; U g = 3 c o s a)t + — c m . C o i b i e n d p s o n g k h o n g d o i k h i
4J
t r u y e n d i . B i e n d p song t o n g h o p tai t r u n g d i e m ciia d o a n A B
A. A = 8cm
B . A = N/34 + 15N/2 c m
C. A = 15cm
D . 2cm.
(' t
Chnyto 3
N e i l d i e m cuoi d a y de t u d o t h i no k h o n g c h i u l u c keo lai ciia gia d o hay
S<^NG
ciia phan d a y sau no, nen n o d i c h chuyen v u g t qua bien d p b i n h t h u o n g
DCTNG
ciia song. D i e m cuoi ciia d a y keo day len phi'a tren v a l u c keo nay l a m
A. K I E N THLTC C a B A N .
phat sinh m o t x u n g phan xa d o n g pha v o i x u n g t o i .
Vay, khi phan xa tren vat can tu do, song phan xa luon ciing pha vai song
1 . S\f p h a n xa cua song.
K h i song gap m o t v a t can hoac d i e m cuoi cua m o i t r u o n g c6 song t r u y e n
toi t h i i t nhat m o t p h a n cua song b j p h a n xa lai.
tai a diem phan xa.
, , ;
2. S o n g d u n g
a) D i n h n g h i a : Song truyen tren sai day dan hoi trong truang hap xuai
a) Phan xa ciia song t r e n vat can co d i n h
Thuc n g h i e m cho thay, k h i phan xa tren vat can co d j n h bien d a n g cua
day b i dao chieu.
hi^
cdc nut vd cdc bunggoi la song ditng.
Song d u n g d u g c h i n h thanh la ket qua ciia s u giao thoa song t o i va song
p h a n xa. N h i r n g d i e m tren day, tai d o cac song triet tieu nhau t h i k h o n g
Tqi sao khi phan xa tren vat can codinh, bien dang bj dao chieu?
dao d o n g va d u g c ggi la nut. N h u n g d i e m tai d o cac song d o n g p h a v a i
Q
nhau t h i dao d o n g v o i bien d p cue dai va d u g c g g i la bung.
N h u n g nut va bung xen ke, each deu nhau.
a)
b) G i a i t h i c h s u tao t h a n h song d i m g t r e n day.
Xet dao d o n g ciia m o t phan
tu tai diem M tren day each
•
Song to^i
dau CO djnh B mot khoang
b)
;<_
M B = d. Gia s u vao thai diem
K h i bien d a n g (xung) t o i dau Q, no tac d u n g m o t l u c h i r o n g l e n v a o gia
d o ( t u o n g ) . Theo d i n h luat 3 n i u t o n , gia d a tac d u n g m g t l u c bang va
t, song toi deh B va truyen
d --.
->
M
toi do mgt dao dong co
n g u o c chieu vao day. Phan luc nay sinh ra m o t x u n g tai gia d a . X u n g nay
phuong trinh dao dong la: Ug = A cos cot = Acos27ift
t r u y e n tren day theo chieu ngugc v a i x u n g t o i . T r o n g p h a n xa nay, phai
Song phan xg
Chgn goc tga do tai B, chieu duong la chieu tu B den M .
CO m o t n u t tai gia d o , v i day bj g i u co d j n h tai do. X u n g t o i va x u n g phan
xa trai dau nhau, de c h i i n g triet tieu Ian nhau tai d i e m d o .
V a y , khi phan xa tren vat can codinh,
song toi a diem phan
song phan xa luon
ngUQC pha
vai
xa.
h) Phan xa cua song tren vat can t y d o
27td^
P h u o n g t r i n h song t o i tai M : u ^ , = A cos 0)t + -
P h u o n g t r i n h song p h a n xa tai B: u g = - A cos (cot) = A c o s ( 2 7 t f t - 7i)
P h u o n g t r i n h song p h a n xa tai M : u ^ = A cos 27tft - 7 1
T h u c n g h i e m cho thay, k h i phan xa tren vat can t u d o bien d a n g cua day
k h o n g b j dao chieu.
Tai sao khi phan xa tren vat can tu do, bii^'n dang khong bi dao chieu?
A
•
Q
N h u vay, tai M dong thoi nhan dugc hai dao dgng cung phuong, cimg
tan so. D o do, dao dgng tai M la tong hgp ciia hai dao dgng do song toi
va song phan xa truyen den:
u = U]^ + U M =ACOS
A
b)
27td^
u = 2Acos
27ld
X
7
C
27lft + —
+ — cos
2
+ Acos
B i e n d o s o n g tai M : A^^ = 2 A cos
doan day thi nang l u g n g lai truyen qua lai t u dau nay toi dau kia, dong thai
27:d
CO su chuyen doi qua lai giua dong nang va the nang. V i the^ k h i xet ve su
bao toan nang lugng, doan day tuong d u o n g con ISc 16 xo.
Bien do nay p h u thuoc i
N e u tchoang each d =
Sv dao dgng cua cac diem tren day khi c6 song dirng
thi bien do dao d o n g tai M bang 0, tai M c6
Tren doan day, t r o n g dieu kien l i t u o n g , cac n u t hoan toan d u n g yen, cac
d i e m con lai van dao dong v o i van toe dao d o n g (can phan biet dugc toe
m o t nut.
+
do dao d o n g cua phan tir m o i t r u o n g v o i toe dot t r u y e n song).
N e u khoang each d =
k + — — thi bien do dao d o n g tai do dat gia t r i
2/ 2
T
T h a i gian giua hai Ian sgi day d u o i thang lien tie'p la — ( K h i day d u o i
eye dai, a dcS c6 m o t b u n g song,
thang, l i do cua b u n g = 0. T h o i gian giira hai Ian lien tie'p l i do b u n g song
V
c) Khoang each giua nut va bung trong song dung
-
H a l b u n g hen tie'p each nhau m o t khoang bang
i
c. T i n h tuan hoan ciia song dirng
H a l nut Hen tie'p each nhau mot khoang bang
-
= 0 la m o t nira chu ki).
*
T i n h tuan hoan theo khong gian:
Khoang each giua b u n g scSng va nut song lien ke bang
^ ,,
Bien do cua phan t u vat chat tai mot diem khi CO song d u n g :
^
Af^^ = 2 A cos
-
^
^.
3. Dac diem ciia song dung
a. S u truyen nang lugng trong song dung
Tai sao goi la song dung? Co phai do nang l u g n g k h o n g t r u y e n di ma
d u n g lai? Neu d u n g lai thi d u n g lai o dau? N e u c6 t r u y e n d i t h i truyen
n h u the nao? Ta hay xem xet van de nay.
Nut luon ctiing yen nen no khong thuc hien cong. D o do nang l u g n g
k h o n g t r u y e n qua d u g c nut. Burij^ khong bien dang, sue cang day tai
b u n g bang 0, ncn hun;^ cung k h o n g thuc hien cong. D o do, nang l u g n g
cung k h o n g t r u y e n d u g c qua bung. N h u vay, nang l u g n g ciia m o i doan
day dai hang 1/4 buoc song c6 m o t dau la nut, dau kia la bung thi k h o n g
d o i . N o i each khac, nang l u g n g " d u n g " trong m o i doan day nhu vay.
N a n g l u g n g cua m o i doan day la khong d o i , k h o n g c6 su t r u y e n nang
l u g n g t u doan day nay sang doan day kia.
T r o n g m o i doan day c6 su truyen nang l u g n g khong? Co s u bien d o i t u
d o n g nang sang the nang khong? Co su t r u y e n nang l u g n g t u d i e m nay
sang d i e m khac cua m o i doan day ma ta xet. N e u t r o n g 1/4 chu k i , nang
l u g n g t r u y e n tir trai sang phai v i d u t u niit tai bung t h i t r o n g 1/4 chu k i
27id
-
; _
n
+—
X
2
Vay, ta coi bien do ciia phan t u moi truong dao dgng dieu hoa v o i chu k i X.
T r o n g t r u o n g h g p chi xet rieng bien do, no c6 the c6 gia t r i am hoac
d u o n g , n h u n g khi xet chung v o i p h u o n g t r i n h song d u n g thi bien do
luon d u o n g .
K h i chi quan tam toi bien do, ta d i m g thuat n g u : " D p l^ch pha bien dp"
dao ctgng t r o n g song d u n g . N h u vay, tren sgi day dan hoi dang c6 song
d u n g on d i n h each nhau m o t khoang d thi dp l^ch pha bien dp la:
2nd
Acp - •
X
Cong t h i i c tren se rat tien Igi trong viec tinh bien do tai m o t d i e m tren
day dang c6 song d i r n g k h i biet khoang each tir no t o i d i e m nut hoac
bung. K h i xet tai dp lech pha bien dp nay ta k h o n g can quan tam toi som
pha hay tre pha v i dieu ta quan tam la do Ion ciia bien do dao dgng.
T i n h tuan hoan theo thai gian:
P h u o n g t r i n h song d i r n g tai m o t d i e m :
u = 2 A cos
27td
n
'27rft-^'
+ — cos
tie'p theo nang l u g n g truyen t u phai sang trai, t u bung toi nut.
X
2
D u a vao p h u o n g t r i n h tren c6 the nhan xet rang : Cac diem tren spi day
Tom lai: Nang l u g n g " d u n g " trong moi doan day dai 1/4 buoc song c6 mot
dan hoi k h i c6 song dimg chi c6 the C U N G P H A hoac N G l / O C P H A .
dau la nut, mot dau la bung. Nang lugng khong truyen ra khoi doan day
Xet hai d i e m M , N tren sgi day dang c6 song d i r n g on d j n h v o i p h u o n g
cung n h u khong truyen vao doan day qua nut va bung. Mat khac, trong m o i
t r i n h Ian l u o t la
Cty TNHH MTV
UM
=2ACOS
2nd
cos ' 2 7 r f t - ^ '
DWH
Khang Vift
duong.
Vay, chiing dao dpng ciing pha.
Ujvj = 2 A cos
+ K h i 2 A cos
^2nd
COS
27rdi
n
+ —
Cac diem thupc hai bo song lien tiep
27rft-^
2
2 A cos
27ld, _^7t"
NimN2 CO bien dp mang dau duong va
bo song N2nNi c6 bien dp mang dau am.
> 0 thi M , N cung pha dao
2
Tir do M va P dao dpng ngupc pha.
dong nghia la phuang trinh bien do mang cung 1 dau (tuc ciing am hoac
4. Dieu ki^n de c6 song dimg tren spi day dai /:
cimg duong) thi chiing dao dong cung pha.
*
+ Khi
2 A cos
27tdi
n
- +—
2 A cos
27id
< 0 thi M , N ngugc pha
(keN*)
So bung song = so bo song = k
So nut song = k + 1
X
2
dao dgng nghla la bieu thuc bien dp trai dau nhau.
-
Hai dau la nut song: 1 =
* Mpt dau la niit song con mpt dau la bung song: 1 = (2k + 1 ) ^ (k € N)
V i tri cac diem dao dpng cung pha, ngupc pha
+ Cac diem doi xiing qua mpt bung thi dong pha (do'i xiing vol nhau qua
duong thSng di qua bung song va vuong goc voi phuang truyen song).
Cac diem doi xung vai nhau qua mpt niit thi dao dpng ngupc pha.
So bo song nguyen = k
So bung song = so nut song = k + 1
B. C A C D A N G B A I T A P
D a n g 1: X a c djnh s o nut s o n g , b u n g s o n g , bo s o n g
m v i DU
MAU
rV
Vi d u 1: (Trich de t h i t h u chuyen Ha Tinh Ian 1 nam 2013)Tren mpt spi
day CO song dirng voi buoc song la A, tren day quan sat thay 4 bung
song. Khoang each giua niit song thu 2 den bung song thu 4 la
A. 1/2
B.2.A.
C.5A/4.
D. A.
^hdn tick vd huong dan gidi
Theo bai ra ta c6 hinh ve:
M , P do'i xung qua bung B nen ciing pha
dao dpng. De thay phuong trinh bien dp cua
M va P cimg dau. Suy ra, M va P dao dpng
ciing pha.
M , Q doi xung qua niit N nen ngupc pha
dao dpng. De thay phuong trinh bien dp ciia
M va Q ngupc dau nhau. Suy ra M va Q dao
1
V
^
1
y
-A
B
dpng ngupc pha.
Cac diem thupc cimg mpt bo song (khoang giiia hai nut lien tiep) thi dao
Ta xem A la nut thir nhat vi the khoang each giira niit song t h u 2 deh bung
dpng cimg pha v i tai do phuong trinh bien dp khong doi dau. Cac diem
song thu 4 la: A. + — = —
4
4
nam 6 hai phia ciia mpt nut thi dao dpng ngupc pha v i tai do phuang
trinh bien dp doi dau khi qua nut.
Mpi diem thupc bo song NimN2 c6 phuang trinh bien dp mang cimg dai'
Chpn dap an C
Bi quyei on luy^n thi dai hgc dat
diem
lot
da
Vat It, tap 1-Le
Van Vinh
Cty TNHH MTV DWH
Vl dM 2: (Trich de t h i t h u chuyen Ha T i n h Ian 2 nam 2013) Song truyen
tren sgi day dai / vai budc song X, de c6 song dung tren sgi day vai mpt
dau day c6' djnh va mpt dau day tu do thi
>
<
41
B. l = k A / 2 v 6 i k = l , 2 , 3 , . . .
vai k = 0, 1, 2,...
A.A =
2k + l
1
D. l = {2k + l)A vol k = 0,1,2,...
vol k = 0,1, 2,...
C.A =
k + 1/2
<PMn tick v>d hu&ng ddn gidi
Chieu dai spi day mpt dau co djnh,
T u ( l ) v a (2) taco: ^
= ^ => f = | = ^
Khang Viet
= lOHz
Sau day la hinh ve bieu dien qua trinh tren:
Niit
mpt dau t u do de c6 song dung la:
Chpn dap an A
Nh^n xet: denha cong thiec tren doi khi de dang
nhimg rat de nham Ian. Khi su dung hinh ve
chung ta se c6 dugc cong thitc tren nhanh va
chinh xdc horn nhieii, sau nay khi giai cdc bai
todn ve tinh nut, bung, bo song hay chieu dai
tuang ung tht sie dung hinh ve Id tot nhat
Tu hinh ve ta c6 chieu dai day thoa man:
Chpn dap an A
Vl dM 4 : (Trich de thi thu chuyen dai hpc Vinh Ian 3 nam 2012)
Mpt spi day dan hoi dai 60cm, toe dp truyen song tren day 8m/s, treo lo limg
tren mpt can rung. Can dao dpng theo phuong ngang voi tan so f thay doi
tu 80Hz deh 120Hz. Trong qua trinh thay doi tan s6^ c6 bao nhieu gia tri tan
so CO the tao song dung tren day?
= k - + - = {2k + l ) - = > X = - ; — 7
2 4 ^
U
2k + l
v a i k = 0; 1;2...
A. 15.
Byng
Vl dv 3: (Trich de thi thu chuyen d?i hgc vinh Ian 3 nam 2012)
B. 8.
C.7.
D.6.
i'hdn tick m huang ddn gidi
Chieu dai day thoa man:
Mpt spi day dan hoi AB hai dau c6' dinh dupe kich thich dao dpng voi
tan so 20Hz thi tren day c6 song dung on dinh vai 3 niit song (khong
tinh hai nut 6 A va B). De tren day c6 song dung voi 2 byng song thi tan
so' dao dpng ciia spi day la
A. 10 Hz.
B. 12Hz.
C.40HZ.
D. 50 Hz.
^hdn tick vd huang ddn gidi
Theo bai ra: khi tan so f = 20Hz thi song dimg c6 5 nut ke ca A va B nen c6 4
bo song. Vay chieu dai day thoa man: \ k^ = 4.^-2X
=^
(1)
Khi tan s o / thi song dung c6 2 byng song nen c6 2 bo song. Vay chieu dai
day luc nay: 1 =
k'y =
2.j
^X' = ^
(2)
Theo de bai: 80Hz < f < 120Hz o 80 < — (2k +1) < 120
11,5 < k < 17,5 => k = 17 -12 +1 = 6
_ V a y CO 6 gia tri cua tan so cho song dung tren day. Chpn dap an D
Vl du 5: Trong thi nghi^m ve song dimg, tren mpt spi day dan hoi dai
1,5 m voi hai dau co dinh, nguoi ta quan sat thay ngoai hai dau day
CO djnh con co n a m diem khac tren day khong dao dpng. Biet khoang
thoi gian giua ba Ian lien tiep voi spi day duoi thMng la 0,1s. V?n toe
truyen song tren day la
. _ A . 5m/s.
B. 4m/s.
C. lOm/s.
D. 6 m/s.
Biquyet on luyjn thi dai hoc dat diem td'i da Vat It, tap 1 - L c Van Vitih
p
B A I TAP VAN DgNG:
C a u 1: M o t sgi day A B = 50cm tree l o l u n g dau A co d j n h , dau B dao d p n g
'Phdn tich vd hiiong ddn gidi
voi tan so 50Hz thi tren day co 12 bo scSng nguyen. K h i do d i e m N each A
m o t doan 20em la b u n g hay n u t song t h u may ke t u A
A. b u n g song t h u 6
C. b y n g song t h u 5
1 = k - + - = k + l- U
2
4
2
CO tat ca 7 n u t song => c6 6 bung song => c6 6 bo song =i> k = 6
= 3 ^ => v = = - ^ ^ ^ = 5 m / s
f
3
3
—
2013)
B la d i e m b u n g thi
d i e m nut.
^kdn
tich vd himng dan gidi
A . (sai) v i d i e m tren day each dau A
mot
doan
bSng nua
buoc
song
k h o n g phai la b u n g song ma la nut
song.
B. (sai) v i d i e m tren day each dau A
m p t doan bSng m o t phan t u buoc
song k h o n g phai la n u t song ma la
b u n g song.
C. (sai) V I d i e m tren day each dau B
m p t doan b3ng ba phan t u buoc
song k h o n g phai la b u n g song ma
la n i i t song.
D . ( d u n g ) V ! d i e m tren day each dau B
m o t doan bSng mot phan t u buoc
song la d i e m nut.
H i n h ve tren se cho cai n h i n cu the hon
Niit
'
,;
,
12 + 2
=Y
= - => A N = 5 - = k - => k = 5 => tai N la n u t t h u 6.
,
C a u 2: M o t spi day A B dai 50em, van toe t r u y e n song tren day la 5m/s, dau A
dao d o n g v o i tan so 80Hz. Tren day co song d u n g hay khong? so'byng song
khi do la :
B. diem tren day each dau A mot doan bSng mot phan t u buoc song la diem nut.
D . d i e m tren day each dau B m p t doan bang m o t phan t u b u o c song la
-
Chpn dap an B
A. diem tren day each dau A mot doan b^ng nua buoc song la diem bung.
C. diem tren day each dau B mot doan bang ba phan t u buoc song la diem bung
^,"
De bie't d u p e N la n i i t song hay b u n g song t h u may ta xet t i so:
C h p n dap an A
K h i CO song d u n g tren spi day dan hoi AB v o i dau A la d i e m n u t va dau
r.
Chieu dai sgi day mot dau t u do, mot dau co d j n h khi co song d u n g thoa man:
Ngoai hai dau day co d j n h con c6 nam diem khac tren day k h o n g dao
V l d u 6: ( T r i c h de t h i t h u chuyen Ha TTnh Ian 2 nam
tich vd hu&ng ddn gidi
f
;
Tren day CO 12 bo song nguyen nen k = 12.
At = T = 0 , ] s = > f = 10Hz
Chieu dai soi day thoa man: 1 = k - = 6 . - =
^
2
2
D . n u t song t h u 5
^hdn
Khoang t h a i gian giua ba Ian lien tiep soi day d u o i thang la m o t chu ky:
dong
B. nut song t h u 6
A . Co, c616 b u n g song.
f
B. Co, co 21 b u n g song.
C. K h o n g .
D . Co, CO 15 b u n g song.
^hdn
tich vd huang ddn gidi
Bai toan chua cho bie't song d i r n g tao ra trong t r u o n g h o p nao nen ta xet ca
hai t r u o n g h p p :
i »
•
T r u o n g h p p 1: H a i dau day e o d m h .
Chieu dai day thoa man: l = k - = k — ^ k = — = ^-^'^-^^ = 1 6 e Z
2
2f
V
5
Suy ra co song d u n g xay ra u n g v o i 16 b u n g song
D o t r u o n g h p p 1 thoa m a n nen ta k h o n g xet t r u o n g h p p 2 nira. T r u o n g hpp
2 la song d u n g tao ra tren day m o t dau co d i n h , m p t dau t u d o .
C h p n dap an A
C a u 3 : M o t spi day dan hoi dai 130cm, co dau A co' d j n h , dau B t u do dao
d o n g v o i tan 50 H z , van toe truyen song tren day la 20 m/s. Tren day co bao
nhieu n i i t va b u n g song:
A. CO 6 n u t song va 6 b u n g song.
B. co 7 n i i t song va 6 b u n g song.
C. CO 7 n u t song va 7 b u n g song.
D . co 6 n u t song va 7 b u n g song.
^hdn
tich vd huang ddn gidi
Nhan xet: song d i r n g tao ra b o i m p t dau t u do, m p t d a u co d i n h
243
=> so niit = so bung = so bo + 1. V i the loai dap so B va D (do bung, nut
khong bang nhau)
Chieu dai soi day thoa man:
1^
21f 1 2.1,3.50 1
k + - -=> k=
1 = k—+ — = k + 2 2f
V
2
2 4
2
20
So' niit = so'bung = so bo + 1 = 6 + 1 = 7.
Cau 4 : Soi day AB = 35cm voi dau B tu do. Tao ra tai A mot dao dong ngang
CO tan so f. Van toe truyen song la 3m/s, muon c6 10 bung song thi tan so
dao dgng phai la bao nhieu ?
B.41,4Hz.
C. 40,7Hz
D. 74,lHz
l U (, 1 ^ V
k+ k+ Chieu dai soi day thoa man:
2?
2
2)
Muon song dung tren day c6 10 bung song => phai c6 9 bo song => k = 9
f= k + 2 21
^ 1^
9+ -
3
2.0,35
= 40,7Hz
D^ng 2: Xac d j n h dieu k i f n de c6 song duTng t r e n d a y
Vl d u 1 : Soi day dan hoi AB, dau A gan voi can rung c6 tan so f, dau B
dugc giu CO dinh. f i va ii la hai tan so lien tiep de tao ra song dung tren
soi day. Tim tan so nho nhat de tao ra duoc song dimg tren sgi day.
B f . =^ = i ± i
21
'min
ijA
2
'min
21
D- mi =^
^hdn tick m hu&ngf danngidi
Z
,
V
fi
V
/ = k — = 1.
2f
(k + l ) - k
V
=
f2-fl
V
f m i n — T T — f9 —i^
21
Vl dM 2: Sgi day dan hoi AB, dau A gan vai can rung c6 tan so f, dau B t u
do. fi va h la hai tan so lien tiep de tao ra song dirng tren sgi day. Tim tan
so nho nha't de tao ra dugc song dung tren sgi day.
. ,j j • ,
A. W
= — = f2 - fi
B f . =^ = i ± i
c
=^ = i z i
D-fmin=^ = f2-fl
f .
41
Zt
Dieu kien de c6 song dung:
(1)
^ ^
Hai tan so fi, k la hai tan so lien tiep de tao ra song dimg ung voi so bo
song Ian lugt la k va k + 1
, Ta CO :
—
41.
= — - — =
2k + l
2(k + l ) + l
A p dung tinh chat: — = — =
b
d b-d
4/
2k + l
^2
2k + 3
_ f 2 - f i
f,-fi
_
(2k + 3 ) - 2 k - l
2
Tan so nho nha't de tao song dirng ung vai k = 1.
= f2-fl
Z/
- I/.
'Phdn tick va hu&ng dan gidi
_
(1)
K
Hai tan so fi, h la hai tan so lien tiep de tao ra song dirng ung voi so bo
^
k+ 1
f2-fl
Tan so'nho nha't de tao song dung ling voi k = 1.
2
Dieu k i f n de c6 song dung: AB = / = k.-^ = k.-;^ ^ ^ =
song Ian lugt la k va k + 1
k
_
AB = / = (2k + l ) . - = (2k + l ) . — — = — ^
^
^4 ^
^ 4f
41 2k + l
ChQH dap an A
C f . = X=
21
h
Chpn dap an A
Chpn dap an C
A. 78,4Hz
V _ f l _
/ = ( 2 k + 1 ) — = (2.0 + 1).
^
Uf
^
V
f
-
V - ^ 2 - f l
^ 4f. i n
m
Chpn dap an C
Cach tinh sau van cho dap an tucmg tu:
Chieu dai sgi day mot dau tu do, mot dau codinh khi c6 song dimg thoa man:
f2
A p dyng tinh chat:
b
a -c
—=
7
a b-d
l=k—+—= k + 2 4
2,
V
Dat: fk = k + 2 2I
k+2
V
k+ 2 21
(1)
94=;
f
fk+l -
I
f
1
^ V
2
J 2T"
k + -- + l
kH
1^ 2
Dau t u d o (dau h o cua ong) la b u n g song.
V
T u (1) va (2) ta c6: k+]-f\,= —
=k—+—=
2
4
Tan so n h o nhat de c6 song d u n g tren day u n g v o l k = 0.
f = k . l
_ V _ fk+uJi—. Suy ra dap an van la C
C
f =
2)
Bung
=> Chieu d a i o n g day thoa man:
' 2y 21 "^21
2
l
A.
1^ v
k + 2y 2 ~ l
2 2?
k + -
v
21"
Tan so'nho nhat de co song d u n g trong o n g u n g v o i k = 0.
340
V l d u 3 : M o t soi day d a n h o i d u o c treo thSng d u n g vao m o t d i e m co d i n h ,
Al
dau con lai tha t u d o . N g u o i ta tao ra song d u n g tren day v o i tan so nho
4.0,85
= 100Hz. Chpn dap an D
Nut
nhat la ^ . Phai tang tan so them m o t l u g n g nho nhat la bao nhieu de lai
CO song d u n g tren day?
A. Af = ^
ra trong ong mot dau kin, mot dau ha giong truang hap song dimg tren sai day
B. Af = 2 ^
3^
C. Af =
, T»}/ nhien, ncu dcy chung ta sc thai/ dugc trifdng hap giao thoa cua song dimg tao
j
mot dau CO djnh, mot dau tie do. VI thechi can dp dung cong thitc tinh tan sonho
D.Af = 4^
nhat cua trmrng hap nay Id co ngay dap dn nhanh nhat.
^hdn tick pd hiccmg dan gidi
v i de cho van toe truyen song va chieu dai day nen ta co
Chieu dai soi day m o t dau t u do, mot dau co dinh k h i co song d u n g thoa man:
k.11
l = k - + - =
2
4
V
k + -
3; 2i
2)
f
=
'min
2) 2\
Tan so n h o nhat de co song d u n g u n g v o i k = 0:
V
2
21
41
m
(1)
.
2) 21
3v
B A I T A P VAN DUNG:
2
T~ = r^-
Vay phai tang t h e m Af = 2^, t h i tren day lai c6 song d i m g .
^
V i d u 4 : N g u o i ta tao song d u n g t r o n g m o t cai o n g m o t d a u k i n m o t dau
ho d a i 0,85m chua day k h o n g k h i 6 dieu k i ^ n t h u o n g (van toe a m la
340m/s). H o i tan so n h o nhat de co song d u n g t r o n g o n g la bao nhieu?
D . f = 100Hz
_
^hdn tick m hu&ng d&n gidi
T r u o n g h o p nay giong t r u o n g h o p song d u n g tao ra b o i sgi day m o t dau co
d i n h , m o t d a u t u do.
D a u CO d j n h (dau k i n cua ong) la n u t .
D . 60 m/s
2f
f
V
K h i f, va (2 la hai tan so lien tiep fj < £3 t h i k j va k j la 2 so
n g u y e n lien tiep: k 2 = k| + 1
C h p n dap an B
Cf-200Hz
C. 32 m/s
. D i e u kien de co song d u n g tren day v o i hai dau day la m i t song (hai d a u co
k
21
djnh): l = k - = k —
— = — = const
41
B . f = 75Hz
B. 2 4 m / s
^hdn tick m hu&ng dan gidi
_ 41 _ .
= ^ = 3=>f, = 3 ^ = ^ A f = f i - ^ - 2 ^ .
A. f - 50 H z
4.0.H5
A. 4 8 m / s
3v
( l ) v a (2)=> f
:
60 H z . Xac d j n h toe do truyen song tren day biet hai dau day deu la n u t song?
(2)
"IT
41
= 100Hz
C a u 1: M p t soi day dai 1 = 1,2 m co song d u n g v o i 2 tan so lien tiep la 40 H z va
Tan so tiep d o de sg'i day lai co song d u n g u n g v o i k = 1
V _
340
—=
N h u vay ta co: — = —
ki
k2
1 _ f2
ki+1
kj
,
,
. . . .
k
21
Van toe t r u y e n song tren day: - = — ^
f
V
40
60
k,
ki+1
21fi
2.1,2.40
,
v= — - = —
= 48 m/s.
kj
2
C h p n dap an A
Tuy nhien, ne'u ndm dugc cong thitc a vi du 1 thi ket qua thu diegc se nhanh han.
Tan so nho nha't de co song d u n g tren day (hai dau co d j n h ) :
^min=Y^=h-h-^^ = 2/(f2
- f i ) = 2.1,2(60 - 40) = 4 8 m / s
247
C a u 2: M o t s(?i day dan h o i c6 m o t dau co djnh, m p t dau t u do, tren day c6
song d u n g v 6 i hai tan so Hen tiep la 30Hz, 70Hz. T a n so n h o nhat de c6
song d u n g tren day la
A. f„.„ = 30Hz
B. f^,„ = lOOHz
^hdn
C. f„i„ = 20Hz
D . f^,„ = 70Hz
tich \pd hitomg ddngidi
,\
Tan so nho nhat de c6 song d u n g tren day (mot dau t u do, mpt dau co dinh)
f n . n = ^ ^ =^
C a u 3 : M p t spi d a y d a n h o i dupe treo thSng d u n g vac m p t d i e m . N g u o i ta
cho d i e m nay dao dpng, tao ra song d u n g tren day v o i tan so nho nhat la f i .
De lai co song d u n g , phai tang tan so'toi thieu den fi. T i so j - bang:
'2
B. 1/3.
'Phdn
C. 4.
tich ra huang
I f ch p h a b i e n d o d e g i a i t o a n
A. K I E N THU'C C A N NAM:
1. Bien d p ciia p h a n t u m o i t r u o n g co p h u o n g t r i n h :
AM
=2ACOS
27id
2. H a i d i e m tren day d a n h o i dang co song d u n g o n d j n h each n h a u khoang
d t h i d p I#ch p h a b i e n d p dao d p n g la: Aep = — ^
= 20Hz.ChpnC
A. 1/4.
p a n g 3 : U'ng d u n g k h a i n i e m
D. 3.
dan gidi
Spi day dan hoi d u p e treo t h i n g d u n g vao m p t d i e m ta hieu rang d i e m con
K h i s u d u n g khai niem ve d p lech pha bien d p de t i n h bien d p k h o n g can
quan t a m t o i khai n i ^ m som pha hay tre pha.
B. B A I T A P V A N D U N G :
Vl d u 1: Tren m p t spi day d a n h o i dang co song d u n g o n d m h , B la m p t
b u n g song, bien d p dao d p n g tai b u n g la A. D i e m M each B m p t doan
d i i n g bang —. T i n h bien d p dao d p n g tai M .
lai d u p e tha t u do v l the'trong t r u o n g hop nay chieu d a i spi day thoa man:
A. A M = 2 A
V
f,
1^ X
1 =k -+- = k + k + 2 4 l
2j 2~ l
2 2f
,
f,
2
2
3
27rMB
>
I
C h p n dap an B
C a u 4 : C h o o n g sao co m p t dau bjt k i n va m p t dau de ho. Biet rang o n g sac
phat ra a m to nhat u n g v a i hai gia t r j tan so cua h a i hpa a m lien tiep la
lOOHz va 280Hz. Tan so am nho nhat k h i ong sao phat ra a m to nha't bang
A. 50 H z .
B.48HZ
^hdn
C. 75 H z .
tich vd huang
D . 90 H z .
dan gidi
t r u o n g h p p song d u n g ciia spi day m p t dau t u do, m p t d a u co'djnh.
-fk
V l the: Li„ =
—- =
C h p n d a p an D
AV3
ddn gidi
280-100
= 90Hz
2.^
3
X
2n
T u h i n h ve ta co:
A M = AR.COS
B-
A
2
Bien d p dao d p n g t r o n g song d u n g co the
am hoac d u a n g hoac bang k h o n g tuy nhien
trong g i a i han ciia chucmg t r l n h , ta chi x e m
Day la t r u o n g h o p tao song d u n g cua song a m . T r u o n g h p p nay giong
,r . u - c
D- A M =
D p l^ch pha bien d p dao d p n g giira M va B:
f , = ^ ( k = 0)
f=
C. A M = A
A
bien d p la d u o n g v i the A ^ = —
VlC h p n 2: Tren B p t spi day dan h o i dang co song d u n g o n d i n h , N la m p t
d u dap an m
nut song, bien d p dao d p n g tai b u n g la A. D i e m M each N m p t doan
d i i n g hang —. T i n h bien d p dao d p n g tai M .
*5
A. A M = 2 A
B. A M = ^
C
A M = A
D. A M =
AV3
249
Cty TNHH MTV DWIi
Toe dp cue dai cua d i e m bung: v max = M A B = ^ . A B = - ^ . 8
0,08
X
Acp =
2nd
_ ^"3
X
_ 2n
X
Khang Vi?t
= 628mm/s
C h p n dap an D
3
V l d u 4 : Tren soi day thang CP spng d u n g , khoang each giira m o t n u t va nut
Sit d u n g vec to quay n h u h i n h ve:
thu 4 ben phai no la 15 cm. D p lech pha giua hai d i e m M , N ( M k h o n g
De d a n g thay rang, bien d o dao d o n g
t r i i n g v a i n u t song) tren day each nhau 1,875cm c6 the c6 gia t r j bSng gia
tri nao t r o n g cac gia t r i sau :
tai M bang —^—
A. 7T/8 rad.
'Phan tich
C h p n dap an D
V l d u 3 : (De thi thu T H P T C h u y e n Phan B g i C h a u - N g h ? A n
nam
B. 3TI/4 rad.
Ian 3
2012)
M , N , P la 3 d i e m lien tiep nhau tren mot soi day m a n g song d u n g c6
C. n/2 rad.
hirnng dan gidi
15
Buoc song: /. = — = 7 , 5 ( c m )
D p lech pha giua hai d i e m M , N :
Acpi^nv^
=
cung bien do 4 m m , dao d o n g tai N nguoc pha v o i dao d o n g tai M .
N P = 2 M N = 2cm . C u sau khoang t h o i gian ngSn nhat la 0,04s soi day c6
d ^ n g m p t doan thMng. Toe d p dao d p n g cua phan ti> vat chat tai diem
b u n g k h i qua v j t r i can bang (lay n = 3,14).
A. 375mm/s.
B. 363 mm/s.
C. 3 1 4 m m / s .
D . 628mm/s.
D . n rad.
27rd
27t.
X
1,875
7
1
2
7,5
Vay chpn dap an C
N h a n xet: )icii
vao each ^idi thi ta klwn^^ thay c6 van dc
sal cd, dap an thu
dugc CO twn^ ban dap an dc cho. Lieu ci'nis^ thirc tinh do tech pha ^ifm hai diem
tnvi^; qua tnnh trui/en son;^ cd khdc
cdn^ thur tinh do lech pha twng song dirng
hay khong?. Co mot dieu cac ban nen nha la: tai hai diem tren sai day cd aong
M , N dao d p n g ngupc pha, cung bien dp dap d p n g nen c h i i n g d p i xiing
dimg thi chi cd the la ciing pha hoac ngiegc pha. Cac diem cd bien do ciing dd'u thi
nhau qua mot n i i t song.
Cling
truyen qua dp dung vdo sdng dimg duw.
qua b u n g nen N , P dao d p n g cung pha. Vay M va
M P
27t.l
<=> n
X
= M N
C7;^
+ NP O
271.2
+X
•
TI =
27r.MN
+
nguvc pha. Vi theta khong
271.NP
— - —
cd song
Cdcli gidi tren Id sai hdn vebdn chat vi
the cac ban can lint 1/ khi Idm triic nghiem.
_yay M va N chi co the cung pha hoac ngupc pha. C h p n dap an D .
V i d u 5: (Dai hpc2011):
J
P dao d p n g ngupc pha nhau =t> M P = n
CP:
thi
tlie sir dung cdiig thur tinh do lech pha ciia hai diem trong moi tritvng
N , P cung bien d p v i the N va P d o i x u n g nhau
Ta
pha vd iiginrc Iqi, cac diem cd bien do khdc dd'u
M o t soi day dan hoi cang ngang, dang co song d u n g on d j n h . Tren day, A
Nut
la mot d i e m nut, B la m p t d i e m b u n g gan A nhat, C la t r u n g d i e m cua AB,
X = 6cm
vpi A B = 10 cm. Biet khoang thoi gian ngan nhat giOa hai Ian ma l i d o dao
De t i m bien d p tai b u n g , ta t i m d p l?ch pha bien d p g i u a d i e m b u n g v a i moi
d p n g cua phan t u tai B bang bien dp dao d p n g cua phan t u tai C la 0,2 s.
t f p n g 3 d i e m tren.
Toe dp t r u y e n song tren day la
NP
o
NB
D p l^ch pha bien dp giira N va B: Acp^g - 2n.V a y b u n g spng CP bien d p : A g =
2
6
A. l),25m/s.
3
= 2 A N = 2.4 = 8 m m
CPS
B. 2 m / s .
C. 0,5 m/s.
D . 1 m/s. '
(Phdn tich vd hitcmg ddn gidi
De d a n g t i n h d u p e buoc song: A = 4.AB = 40cm.
D p lech pha bien dp dao d p n g giOa B va C:
Bung B
-
3
T
T h o i gian g i i i a hai Ian day d u p i thSng lien tiep la — = 0,04
T = 0,08s .
'
\(PBC
= 271
Niit A
Bi quye't on luy^n thi dai hgc dat diem tot da Vat U, tap 1 - Le Van Vinh
tmin(uB=Ac^UB=Ac) " V A B ^
>AB
^ *^
Van toe t r u y e n song tren day: v =
Cty TNHH MTV DWH Khang Vi^t
'Phan tich vd hu&ng dan giai
A B 7 2 r ? + ? = ? = 0 ' 2 = > T = 0,8s
AB-
=
8
8
4
= 50cm / s = 0,5m / s
T
Buae song: A B = —
4
p. ^ i
=:>A, = 4 A B = 4.18 = 72cm
0,8
Dp l?ch pha bien d p g i i i a B va M :
ChpndapanC
^..^^
MB
A(p = 2n-
Ta c u n g c6 the t i n h chu k y theo v o n g tron l u p n g giac n h u sau:
T h a i gian ngan nha't de l i d p dao d p n g cua p h a n t u tai B bang bien d p dao
271.12 71 ,
- ^ — ( r a d )
Bien d p dao d p n g tai M :
d p n g tai C u n g v o i goc quet 2.45° = 90".
AM =AB.COS| = 2A.COS| = A
Vay © = - = ^
= 2,57i(rad/s)
t
0,2
^
'
= — = l,25(Hz)
2n
Toe d p cue d a i eiia phan t u dao
-2ACO
-AW
Aco
-•
v = ?..f = 40.1,25 = 5 0 ( c m / s )
d p n g tai M : v^max = « A
T a cung c6 the giai bai toan tren theo phuong trinh song n h u sau
Toe d p eye d a i cua p h a n t u m o i t r u o n g
Ta CO bien d p song d u n g tai m p t d i e m M tren day, each d a u c o ' d i n h A doan
tai b u n g song la: Vg^ax = ' ^ • 2 A
T h a i gian c6 v a n
d la:
Theo bai ra:
2Aco
•
toe n h o h a n VMmax
. 27td
v o i a la bien d p n g u o n song. Ta c6:
A M = 2a sm
A t - 4t
N
O
M
T
T
o-^^^-Bmax = 4 . — = j = 0 , l = ^ T = 0,3s
X
Bien d p song tai d i e m B ^dn = - = 10=>X = 40cm^ la: A B = 2a
Toe d p t r u y e n song tren day: v = | = ^
Bien d p song t ^ i d i e m C:
= 240em/s = 2 , 4 m / s
C h p n dap an D
2K-
( d c ) = — = g = ^ A c = 2a sm
V I d u 7 : T r e n m p t spi d a y d a n h o i chieu d a i / = 80cm, h a i d a u co d j n h v a
8 = 2a.— = A n —
2
^ 2
dang CO song d u n g . Q u a n sat tren day tha'y c6 cac d i e m each deu nhau
n h i i n g k h o a n g 10cm l u o n dao d p n g cung bien d p nhau. So n u t song tren
day la:
Co the coi d i e m B n h u m p t chat d i e m dao d p n g d i e u hoa v o i b i e n d p A B , thi
72
^
t h a i g i a n ngan nha't g i i i a h a i Ian d i e m B c6 l i d p A B — la:
A. 4
B^8
C. 5
D. 7
^hdn tich v>d huang dan gidi
At = - = 0,2=i>T = 0,8s=> v = - = 0 , 5 m / s
4
T
N h a n x e t : khi sit dung do lech pha bien do de gidi bai toan tren thi bai toan tren
khd giohg vai bai toan ve thai gian trong dao dong dieu hoa vi these de hieu va
ngan gon han nhieu so vai each giai khdc.
V I d u 6 : M p t sai day dan h o i cang ngang, dang c6 song d u n g o n dinh. Tren
day, A la mpt diem niit, B la diem b y n g gan A nha't v o i A B = 18cm, M la mpt
d i e m tren day each B m p t khoang 12cm. Biet rang trong m p t chu k y song/
khoang thai gian ma dp Ion van toe dao dpng cua phan t u B nho h a n v a n toe
Gpi M , N , P, Q, R la cac d i e m lien tiep nhau dao d p n g c u n g bien d p .
Theo de eho ta c6: M N = N P = PQ = QR =10 c m
eye dai ciia phan t u M la 0,1s. Toe dp truyen song tren day la:
Theo h i n h ve buoc song:
A . 3,2m/s.
Chieu d a i spi day hai d a u co d i n h thoa man:
B. 5,6m/s.
C. 4,8 m/s.
D . 2,4 m/s.
= M R = 4 M N = 4.10 = 40em = 0,4m
|
. j t ,.i, ;|.
Bi qiiyet
on
luyfit
thi
itai IIQC
dat
diem
td'i da
Vat
It, tap
1 - Le Van
Vinh
, X
,
21 2.0,8
.
1 = k-=> k = — = — — = 4
2
X
0,4
,
Vay tren day c6 4 bo song => c6 5 n u t song. C h p n dap an C
m
B A I TAP VAN DUNG:
C a u 1: Tren day A B co song d u n g voi dau B la mot nut. Song tren day c6 bu6c
song A. Hai diem gan B nhat co bien do dao dong bSng m o t nua bien d g dao
dong cue dai cua song d u n g each nhau mot khoang la:
A. A/3;
B.A/4.
C. A/6;
D . A/12;
:=> Buoc song: A, = M R = 4 M N = 4.15 = 60em
'
V a n toe t r u y e n song tren day: v = A,f = 60.50 = 3000cm / s = 3 0 m / s
C h p n dap an B
C a u 3 : M o t song d u n g tren m g t doan day co budc song bang 30em va bien
'Phdn tich ra hii&ng dan gidi
Ggi C la b u n g gan n u t B nhat n h u hinh ve va M , N la hai d i e m co bien do
dao d o n g b5ng m o t nua bien do dao d g n g cue dai (bien d g dao d o n g cua
d g dao d g n g eiia m g t phan t u each m g t n i i t song m g t doan 5cm co gia t r i la
9 m m . Bien d g A cua b u n g song la:
A. 9V2 ( m m )
d i e m C).
B. 18 ( m m )
C. 9 ( m m )
D . 6V3
(mm)
i'hdn tich i?d hitong dan gidi
A,
Ggi N la n i i t va B la b u n g gan N
T u h i n h ve ta co: c o s C O M = - r - = » => C O M = A,.
nhat, d i e m each n u t 5cm la M .
T u o n g t u ta cijng co:
D g l^eh pha bien d g tai N va M :
CON = C O M = -
271
=> M O N - y
.
„ MN
Ag = 271
= A(PMN
^
I
D o lech pha giua hai d i e m M , N :
5
= 271—
n
=
30
—
3
T u v o n g tron ta thay, l i d g ciia
ACPMN
=27t-
3.27t
In
X
b u n g song (diem B) tai d i e m M la:
"B =
= — ^ = 9 => A = 6 7 3 m m . C h p n dap an D
3
C h p n dap an A
Bhni; each si\ khoan;; thai giaii dqc biet, ta gidi nhanh nhu
min(M->N) -
^A Q
,
2 "'^^
+t
-1
Ac- 2
"
A
sau:
tren day tai m g t t h o i d i e m khong the
1-1
~
ri
A . dao d g n g n g u g c pha.
ihco tilth tiiifii hoan cua song theo khong gian va thai gian ta co:^ <^ ^
Viiy M N =
C a u 4 : Tren m g t sgi day dan hoi dang xay ra song d u n g , hai d i e m rieng b i ^ t
B. dao d g n g l^ch pha —,
3
C. dao d g n g lech pha ^ •
D . dao d g n g ciing pha.
. C(7c7! mn/ nhanh va tirang doi dchicu nen cac ban chi'i y.
C a u 2: Tren mot sgi day cang ngang v o i hai dau co d j n h dang co song d u n g
^hdn
tich vd hu&ng dan gidi
Xet A la n i i t va B la m g t b y n g song gan nhau
K h o n g xet cac d i e m b u n g hoac nut, quan sat thay nhi>ng d i e m co cung bien
nhat n h u h i n h ve.
do va 6 gan nhau nha't thi deu each deu nhau 15cm. Tan so dao d g n g tren
Ggi M , N d o i x u n g v o i nhau qua b u n g B suy ra
day la 50 H / . . Van toe t r u y e n song tren day co gia t r j b5ng
M , N dao d g n g cung pha.
A . 30em/s.
B. 30 m/s
C. 90 cm/s.
D . 45 m/s.
Theo de cho, ta co: M N = N P = PQ = QR =15 c m
Ggi N , Q d o i x u n g v o i nhau qua n i i t A suy ra N ,
Q dao d g n g n g u g c pha.
Hai diem doi xwtg myi nhau qua mot bung thi dao dong cung pha.
Hai diem dot ximg vai nhau qua mot nut thi dao dgng nguac pha.
N e u xet ve pha thi tren m o t soi day dan hoi dang xay ra song d i r n g chi co
S6NG A M
the ciing pha hoac ngugc pha, khong c6 v u o n g pha va cac pha khac.
C h p n dap an C
*
Cau 5: M o t soi day dan hoi cang ngang, dang c6 song d u n g o n d i n h . Tren
1. Song am - cam giac am:
« Song dm la n h u n g dao dong co truyen trong cac m o i t r u o n g k h i , l o n g , ran.
day N la m o t d i e m nut, B la m o t d i e m b u n g gan N nhat, N B = 25 cm, goi C
T r o n g chat long va chat k h i song am la song doc, t r o n g chat ran song am
A \f3
la m o t d i e m tren N B c6 bien do A ^ = — ~ — . K h o a n g each N C la
2
g o m ca song ngang va song doc.
B.
A . ^
3
40
C. 50
4 Phdn loai:
-
D . 40
, , ,
Song am co tan so t u 16Hz den 20000Hz la n h i i n g dm nghe dwoc va
t h u o n g goi la dm thanh.
-
N , B la n u t va b u n g gan nhau nhat => N B = —
4
iJ i
•
Song am co tan s o l o n h o n 20000Hz gpi la sieu dm.
Tai n g u o i k h o n g nghe dupe ha am va sieu am.
Buoc song: X = 4 N B = 4.25 = 100 cm
Theo bai ra: bien do A ^ =
Song am co tan so nho h o n 16Hz goi la ha dm.
N h i r n g am co tan so xac d j n h {iiaxj do thi dao dong cua dm bien thicn tuan
ABV3
hodn) t h i gpi la nhac am, con n h i r n g am co tan so k h o n g xac d i n h (hay do
thi dao dong cua dm bien thien khong tudn hoan) t h i gpi la tap am.
Do lech pha bien do tai B va C la:
,
„ BC 7t
^„
X 100 50
Acp = 271
= — => BC = — =
= —cm
X
3
6
6
2. Cac dac tinh vat li:
'
a. Tan so am: Tan so am bang tan so dao d p n g ciia n g u o n
'
f b. Cubng do dm - Muc cuang do dm:
3
C h p n dap an C
Cau 6: Mot soi day dan hoi cang ngang, dang c6 song dung on djnh. Tren
day A la mot niit, B la diem bung gan A nhat, AB = 14 cm. Bien do tai bung
14
song la 2A. C la mot diem tren day trong khoang AB voi AC = —cm . Bien
3
do dao d o n g tai d i e m C
C u a n g dp am:
j._ P _ W
. ( D o n vj : w/m^)
~ S ~ S.t
P la nang l u p n g dao d p n g am truyen qua dien ti'ch S t r o n g I s (con gpi la
cong suat am); S la dien tich v u o n g goc v o l p h u o n g t r u y e n am.
M u c cuong dp am: la dai l u p n g d u n g de so sanh dp to cua a m v o i c u o n g
dp am chuan, co d o n v j la fcen(B).
AVI
B. A c = AN/3
A . A r =•
C.
AC = A
D.
AC = 2 A
C o n g thuc d j n h nghla m u c cuong dp am la:
v i I B = lOdB
^hdn tick vd hu&ng ddn gidi
Theo bai ra: Tren day A la mot nut, B la diem
b u n g gan A nhat nen X = 4.AB = 4.14 = 56 c m
D o lech pha bien dp ciia A va C la:
14
n
V2
6
C h g n dap an C
^0
c. Do thi dao dong am: T o n g h p p quy luat dao d p n g a m ciia tat ca cac hpa
am t r o n g m o t nhac am cho ta do thi dao d p n g am cua nhac am do. D u n g
T u v o n g t r o n ta c6:
7t
.Ta C cong thuc: L(dB) = 10.1g-^ . M u c c u a n g dp a m t h u o n g gap co trj
O
so vao khoang t u 20dB den lOOdB.
AX
56 6
AC=ABCOS
nen k h i d u n g d a n v j dexiben(dB)
AR
2A
cu khao sat do thj dao d p n g am la dao d p n g k i dien t u , n g u y e n ly hoat
-A.
d p n g la bien dao d p n g am thanh dao d p n g di§n.
Cty TNHH MTV DWH
3. Cac dac tinh sinh li:
a. Do cao cua dm: Do cao ciia am la dac tinh sinh ly cua am do tan so' am
quye't dinh, am cao thi c6 tan so Ion hon am tram (thap).
'Phdn tich vd hu&ng dan gidi
A. (diing) v i toe dp truyen am trong moi truong la v = Xi ti le voi tan so am
va buoe song nhung chii y rang to'c dp truyen am trong moi truong
khong phu thupc vao tan so am va buoc song ma chi phu thupe vao moi
truong truyen am.
B. (diing) vi am nghe dupe c6 eiing ban chat vai sieu am va ha am chi khac
nhau ve pho tan so'. Cu the la: am nghe dupe c6 tan so' t u
16 < f < 20000(Hz), ha am eo tan so t u f < 16(Hz) va sieu am c6 tan so tu
f > 20000(Hz).
b. Am sac: A m sac do quy luat bien thien tuan hoan ciia cac hpa am (do thi
dao dong am) quye't dinh.
c. DQ to cua dm:
-
6 mot tan so xac dinh, cuong dp am cang Ion, cho ta cam giac nghe thay
am cang to. Tuy nhien do to cm dm khong ti le v&i thuan voi cuang do am, ma
no con phu thugc vao tan so ciia dm.
-
De am thanh gay ra dupe cam giac am, cuong dp am phai Ian hon mot
C. (sai) vi am sac, dp cao, dp to la nhiing dac trung sinh ly ciia am eon tan
so'la dae trung vat ly ciia am.
gia tri cue tieu nao do gpi la nguang nghe. Ngudmg nghe thay doi theo tan so
cua dm, v6if=
lOOOHz thi ngudmg nghe la Um = L = 10 ' " W/m^ {con goi la
D. (diing) v i song am la cac song eo truyen trong cac moi truong ran, long,
cuong do dm chuan)
-
khi nhung khong truyen dupe trong ehan khong.
Khi cuong dp am len den lOW/m^ (ung voi muc cuong dp am 130dB) doi
Chpn dap an C
voi mpi tan so cua am deu gay cho tai nguoi nghe cam giac nhuc nhoi,
Vi du 2: Song am truyen trong khong khi den tai nguoi. Tai c6 the cam nhan
rat kho chiu. Gia trj nay ciia cuong dp am gpi la nguong dau.
-
dupe am c6 ehu ky bang?
Dp to ciia am phu thudc vao tan so dm vd cuang do dm hay muc cuong do dm.
A. 0,2 s
Dp to ciia am ung voi mot tan so xac dinh boi: AI = I - Imin. Mien nghe duac
1
1
•
—=
= 5Hz => day la song ha am
T 0,2
^
^ •
J. Vai T = 0,5^is = 0,5.10~^s => f = - =
^
= 2.10^Hz => day la song
T 0,5.10"^
^
sieu am
5. Neu nguon phat am la nguon diem phat am c6 dinh hudng theo Yanh
A. Voi T = 0,2s
quat cau thi S la di^n tich chom cau: S = 27i.R.h
voi h = R - R c o s - = R 1-cos—
2
2
1-cos—
2.
!. Voi T = 10ms = 10"^s
V I DU MAU
=
= lOOHz => day la song am ma tai
^
D. Voi T = 0,5ns = 0,5.10'^s => f = - =
^—Q- = 2.10^HZ => day la song
T 0,5.10'^
^
sieu am
'iu-v
Ket luan nao khong dung voi song am?
A. Toe dp truyen am trong moi truong ti 1§ voi tan so'am.
D. Song am la cac song eo truyen trong cac moi truong ran, long, khi.
=> f =
nguoi nghe dupe
Vl du 1: (Trich de thi thu chuyen Ha Lpng - Quang Ninh Ian 1 nam 2013)
C. A m sac, dp cao, dp to, tan so la nhiing dac trung sinh ly eiia am.
D. 5 ns
ngay dap an.
(R la ban kinh mat song cau)
B. A m nghe dupe eo eung ban chat voi sieu am va ha am.
C.lOms
ung voi timg chu ky decho vd sau do so sdnh vai tan soma tai ta nghe duac Id cd
4. Cong suat ciia nguon am (nguon diem): P = S.I = 47cR^.I
Q
B. 0,5^s
Tai ta cam nhan dime dm c6 tan so tit 16Hz den 20000Hz vi the chi can tim tan so
Tai nguoi nghe dupe am c6 muc cuong dp am t u 0 (dB) den 130 (dB).
Cong suat nguon am la: P = 27tR
. , >:
'Phan tich vd hu&ng dan gidi
ciia am nam trong khoang tu Imm den Imax.
-
KffangViit
^ C h p n dap an C
____
1 Vl du 3: Mpt nguon song am dupe dat trong nuoc. Bie't khoang each giiia
mm
hai diem gan nhau nha't dao dpng ciing pha nhau la 2,5m va van toe truyen
am trong nude la 1,8.10'^m/s . Tan so'ciia song am do la
A. 0,6kHz
B. 1,8kHz
C. 0,9kHz
D. 3,2kHz
Bl
lJUyet
on lUy^
tWl UUl ni^U uui
uwm
ivi
uu
vui
n, lUff
I—
cry
'Phdn tich v>d huang dan gidi
D o lech pha cua hai d i e m dao d o n g cung pha: A(p =
=> A d =
1,8.10^
c u o n g d p tai d i e m N each nguon m o t doan 15m la:
A . 130dB
' ? T t =':il'S"
B. 150,5dB
C. 170dB
D . 116,5dB
'Phdn tich vd huong ddn gidi
Theo de ra, ta c6: O M = 100m
,
M u c c u o n g d p a m tai M va N Ian l u p t la:
t r o n g m o t m o i t r u o n g truyen a m dang h u o n g va k h o n g hap t h u am. H a i
d i e m A , B each n g u o n a m Ian luQ-t la R i va R2. Biet c u o n g d o a m tai A gap
= 1 0 . l o g ^ va Lj^ = l 0 . 1 o g ^
Hieu m u c c u o n g dp a m tai A va B la:
16 Ian c u o n g d o a m tai B. T i so — nhan gia t r j nao sau d a y
C. 1/16.
Viet
c u o n g d o a m la 100 dB. Gia su m o i t r u o n g k h o n g h a p t h u a m . M u c
V lC h pu 4 : M o tCn g u o n am dat tai O phat song a m c6 cong suat k h o n g doi
d n dap an
B. 16.
nvvn~Knang
dang h u o n g ve m o i p h u o n g . Tai d i e m M each O m o t doan 100m, m u c
= kin
Tan so cua song am: => f = — = - ^ - ^ — = 900Hz .
A . 4.
Miv
V l d u 6 : N g u o n a m O phat ra m o t a m c6 cong suat P k h o n g d o i , t r u y e n
v a i k e N * => k ^ j ^ = 1 => Ad^^ir, = > = 2=>;^ = 2 m
^
V
iNHii
L M - L N = 1 0 . 1 o g i ^ = 10.log
D . 1/4.
fON^
2
[oMj
= 10. log
2
f 15 ^
llOOy
= -16,5
L N = L M +16,5 = 100+ 16,5 = 116,5dB.
^hdn tich v>d hu&ng ddn gidi
C h p n dap an D
V l m o i t r u o n g phat song a m la dang h u o n g n e n song phat ra d u a l dang
V l d u 7 : Tai m o t d i e m A n a m each xa n g u o n a m N (coi n h u n g u o n diem) 1
h i n h cau nen d i ^ n tich mat cau dugc xac d i n h theo cong thuc: S = 47iR^
khoang N A = 8 m eo m u c c u o n g d p a m la L A = 110 dB. Biet n g u o n g nghe
Ta biet rSng: c u o n g d o a m t i 1# nghjch v o i b i n h p h u o n g khoang each t u
p
p
d i e m can t i n h c u o n g d o a m v o i n g u o n am, cu the la: I = — = — r y
eiia a m d o la L. = 10"
W / m l Xet d i e m B n a m tren d u o n g N A va each N
khoang N B = 10 m . C u o n g d p a m tai B la:
A . 6.10^^ W / m l
I A = 16IB O — = 1 6 —
B. 6,46.10"2 W/m2.
C. 10-2W/m2.
Theo bai ra, ta c6:
D.10--^W/m2.
'Phdn tich m huong ddn gidi
S B = 1 6 S A O 47iR^ = 16.47tR?
W M u c c u o n g d p a m tai A va B Ian l u p t la:
R?
Ri
H i e u m u c c u o n g dp a m tai A va B la:
C h q n dap an A
L A - L B = 1 0 . 1 o g ^ = 10.1og ^
V l d u 5: M o t song a m t r u y e n t r o n g k h o n g k h i . M i i c c u o n g d o a m tai d i e m
M va tai d i e m N Ian l u o t la 50 dB va 100 dB. C u o n g d o a m tai N Ion h o n
p M u c c u o n g d p a m tai B: L g =
c u o n g d o a m tai M .
A . 100000 Ian
B.1000 Ian
C. 50 Ian
M u c c u o n g dQ a m tai M va N Ian l u g t la:
=1,94
- 1 , 9 4 = 1 1 0 - 1 , 9 4 = 108,IdB
L B = l O l o g l ^ = 108,1 ^ I B = 10^°'«llo = 10'°''\lO-'^
= 6,46.10-^
m
^0
O i Q n d a p an B
= 1 0 1 o g i ^ va L M = 1 0 1 o g y i
Vl d u 8 : Ba d i e m O, A , B cimg nam tren mot nua d u o n g thSng xuat phat tiJ' O.
H i f u m u c c u o n g d p a m tai N va M :
L N - L M = 1 0 1 o g i ^ = 1 0 0 - 5 0 = 50 ^
rio^
=10.1og ^
C u o n g d p a m tai B:
D . 5 0 0 0 Ian
'Phdn tich vd huong ddn gidi
C h p n dap an A
= l O l o g — va L g = l O l o g —
Tai O dat m o t nguon diem phat song a m d i n g h u o n g ra k h o n g gian, m o i
^
t r u o n g k h o n g hap thu am. M u c cuong dp a m tai A la 100 dB, tai B la 60 dB.
= 10^ ^ 1 ^ = I O ^ I M •
M u c cuong dp am tai diem M thupc doan A B v o i A M = 4 M B la
_
A. 61,9dB.
B. 72,6dB.
C. 43,6 dB.
D . 70,5 dB.
'Phdn tich vd hit&ng ddn gidi
^hdn tick vd hu&ng ddn gidi
O
A
B
M
M u c c u a n g d p a m tai A va B Ian luqt la:
=10logva
LM=101og
Lg = 1 0 l o g ^
0 ;
LM=101og
I
H i e u m i i c c u o n g d p a m tai A va B:
L A - L B = 1 0 1 o g i ^ = 100-60 = 4 0 = ^ ^ = 10''=f
5
5
0 A + 4 0 B
= 20 + L
RM^401
RA
40n
= 100-10.1og
401^
Chpn dap an B
Vl d u 1 1 : (Trich de thi thu chuyen H a T i n h Ian 2 nam
2013)
M o t n g u o i d u n g each n g u o n a m m o t khoang la x t h i c u o n g d p a m la I .
K h i n g u o i d o tien ra n g u o n xa them m p t doan 30m t h i c u o n g d p a m
g i a m chi con bang 1/4. K h o a n g each x ban d a u la
= 61,9dB.
A . 60m.
J
B. 15m.
C. 7,5m.
D . 30m.
'Phdn tich vd hu&ng ddn gidi
M o t dan loa phat am thanh dSng h u o n g . M u c c u o n g dp a m do
Gpi A la d i e m each n g u o n a m m p t khoang x (x > 0) c6 c u a n g d p a m 1 ^ = I
dupe tai cac d i e m each loa m o t khoang a va 5a Ian l u p t la lOOdB va L. Gia
Gpi B la d i e m each n g u o n a m m p t khoang x +30 c6 c u o n g d p a m Ig = -
trj ciia L la
B.49dB
C. 86dB.
4
D . 25dB.
Ta biet rang c u o n g dp am t i
'Phdn tich vd hic&ng ddn gidi
O
le nghjch v o i b i n h p h u o n g
G p i m u c c u o n g d p am tai d i e m each loa m o t khoang a la L^.
G p i m u c c u o n g d p a m tai d i e m each loa m o t khoang 5a la L^^.
c u o n g d p a m den n g u o n nen
H i ^ u m u c c u o n g d p a m tai hai d i e m do la:
ta c6:
IA=1
khoang each t u d i e m can t i n h
2
IA
= l O . l o g i ^ = lOlogf
- 10.1og25 = 14
[RAJ
= L a - 1 4 = 100-14 = 86dB.
loA,
o - 3 x ^ + 6 0 x + 900 = 0:
Vi d u
Xet d i e m M 6 t r o n g m o i t r u o n g dan h o i c6 song a m t r u y e n qua. M i i c
C. l O L ( B ) .
,
D . lOO.L (B).
X
J
= 4 o x ^ + 6 0 x + 900 = 4x^
x = 30
x = -10(l)
1 2 : Tai O c6 1 n g u o n p h a f a m thanh dSng h u o n g v d i cong suat
k h o n g d o i . M p t n g u o i d i bp t u A den B theo 1 d u d n g thJing va ling nghe
c u o n g d p am tai M la L (B). Ne'u c u o n g dp a m tai d i e m M tang len 100
B. 10.L + 20(dB).
x + 30
^ h p n d a p an D
V l d u 1 0 : (Trich de thi thu chuyen H a T i n h Ian 1 nam 2013)
Ian t h i m u e c u o n g d p a m tai d i e m do bang
{
IR=B
roB^ 2 fx + 30\
C h g n dap an C
A . L + 20 (dB).
bay"
ciianguairade
5
C h p n dap an A
L, -L53
(dB)
L(B) = 1 OL(dB) nhu vay dap an A chua dung. De dang thay ring dap an phdi Id
R«
Mon
L A - L M = 1 0 . 1 o g ^ = 10.1og - M - =10.1og
1M
V ^"^A J
A . lOOdB.
= 101oglOO + 101og
;
B. Vai nhung cau the nay, ne'u cdc ban khong dey den dan vi thi rat de "mac
H i ? u m u c c u o n g d p a m tai A va M
Vi du 9:
1001 M
nhimg decho muc citang do am tai M la Ben (B) vi the can doi ra dB.
R A + 4 R B _ R A + 4 .lOOR A _ 4 0 1 R A
V 5
0
= 101og
C
c h i i y: mi'fc cuang do dm trong cong thiec tren tinh theo dan vi deciben
O M - O A = 4 ( O B - O M ) r ^ 5 0 M = OA + 40Bc:>OM =
LM=LA-10.1og
M'
C
C h p n dap an A
RB=100RA
Theo bai ra: M A = 4 M B
1
M
am thanh t u n g u o n O t h i nghe thay c u o n g d p a m tang t u 41 den 251 r o i
iji i/ui/i'i uji miivn
n i l uiii iw
till
lai c i a m xuonc 41. Ti so
• ^
A.
^
5V21
AO
V
iim
d, - d i
la
^
AB
B. 2N/2T
2
D.
27t,
=(A<PN + A ( p ) — = > A(PN = - - ( d 2
K
In
-di)-A(p
Theo bai ra: d 2 - d j = kX va A ( p = rt
5^/5
Thay vao: A c p ^ = — " ( ^ 2 ) - Acp = - ^ . k ) . - n = (2k - 1 ) 7 1 thuoc so le Ian TT
(Phdn tich vd huang ddn gidi
K
Theo bai ra ta c6: cuong do am tai A va B c6 gia trj bang nhau ket hop v6i
gia thie't de cho la song am truyen d u o i dang song cau (am phat dSrig
h u o n g ) v i the A va B ciing nam tren m o t d u o n g t r o n . Vay d i e m c6 cuong do
am Ion nhat tren doan A B chinh la d i e m H nam tren d u o n g t r u n g true cua
A.
nen tai N song dao d p n g v o i bien do cue tieu. V i bien dp cua hai nguon
bang nhau nen bien d p tai d i e m N bang 0.
Chon dap an A
Vi du 14: (Trich de thi thu chuyen nguyen Trai - Hai Duong Ian 1 nam 2013)
tam giac O A B ( n h u h i n h ve)
M o t am thoa c6 tan so 440 H z (phat ra am la) dat sat m i e n g m o t b i n h tru
Theo bai ra: 1 ^ = I B = 4 1 => O A = O B
d u n g nuoc c6 m u c nuoc each mieng b i n h sao cho am thanh phat ra t u
lH=Imax=25I:
25
4
mieng b i n h la to nhat. H o i can rot them vao b i n h m o t cot nuoc c6 chieu
OA^
OH.
cao toi thieu la bao nhieu t h i am thanh tro nen nho nhat? V a n toe t r u y e n
am t r o n g k h o n g k h i bang 330m/s.
•OH=-OA
5
A. 18,75cm
B. 17,85 cm
Song d u n g tao ra t r o n g b i n h h i n h t r u m o t dau
A O ^ = O H ^ . H A ^ = l ^ . ^
25
4
21AO^
AB^
AO
5721
4
k i n , m o t dau ho c6 chieu dai b i n h thoa man:
l = (2k + l ) ^
AB
C h o n dap an A
A m thanh t r o n g b i n h phat ra t u gia t r j to nha't
I A = 4 I
V I d u 13: (Trich de t h i t h u chuyen H a L p n g - Q u a n g N i n h Ian 1 n a m 2013)
Tai hai d i e m P va Q t r o n g k h o n g k h i c6 hai n g u o n song am cung tan so f,
cijng bien do A , do I f c h pha la TI. Song am t r u y e n t u hai n g u o n am do v o i
buoc song A den d i e m N nam ngoai d u o n g thang PQ, c6 hieu khoang
each den P,Q la k A ( v o i k = 1,2,3...). Coi m o i t r u o n g k h o n g hap t h u am.
K h i do, tai d i e m N
A . hai song giao thoa nhau u n g v o i bien d g cue tieu la A N = 0.
B. hai song giao thoa nhau u n g v o i bien do cue dai la A N = 2 A .
C. hai song giao thoa nhau u n g v o i bien do la A N = A 7 2 .
D. hai song k h o n g giao thoa nhau n h u n g c6 bien dp song la A N =A 0.
'Phdn tich vd huong ddn gidi
H a i song tren thoa m a n dieu kien giao thoa v i the m u o h biet tai m o t diem
dao d o n g v o i bien d p cue dai hay cue tieu ta lam n h u sau:
A p d u n g cong thuc t i n h tong quat ben giao thoa:
D . 27,5 cm
^hdn tich vd huang ddn gidi
T r o n g tam giac v u o n g O H A ta c6:
25
C. 37,5 cm
sang gia trj nho nhat k h i tai mieng b i n h chuyen
t u b u n g song sang n i i t song nghia la m u c nuoc
can do them vao tol thieu phai bang v o i khoang
each g i i i a n u t va b u n g lien tie'p la:
_ V
_
330
f
4 ~ 44f ~ 4.440
= 0,1875m = 18,75cm
C h p n dap an A
Vi d u 15: ( T r i c h de t h i t h u chuyen d a i hpc V i n h Ian 3 n a m 2012)
Cho o n g sao c6 m o t dau bjt k i n va m o t dau de ho. Biet rang ong sao phat
ra am to nhat u n g v o i hai gia t r j tan so cua hai hpa am lien tie'p la 150 H z
va 250 H z . Tan so am nho nhat k h i ong sao phat ra am to nhat bang
A . 50 H z .
B.75HZ.
C. 25 H z .
'Phdn tich vd huang ddn gidi
Chieu dai 6'ng sao thoa man:
Uk^4
= (2k.l)^ = ( 2 k . l ) ^ ^ f = (2k.l)^=>^=^
D . 100 H z .
Bi quyet on luy?n thi dai hgc dat diem tot da Vat It, tap 1 - Le Van
Vinh
'Phdn tich ra huorng ddn gidi
fk=(2k + l) —
Goi I la c u o n g d p a m do m o t chiec ken phat ra tai v i tri do m u c c u o n g d p am.
Theo bai ra ta c6:
C u o n g d p a m do 10 chiec ken phat ra tai v i t r i do se la 101
f,,,=(2(k + l) + l ) ^
Gpi n la so ken de tai v i t r i tren do dupe m u c c u o n g d p a m la 60dB v i t h e t a
c6:
2
0
2
Lio = lOlog
C h p n dap an A
du 16:
Vi
(Trich de thi dai hpc 2012)
|
Tai d i e m O trong m o i t r u o n g d i n g h u o n g , khong hap t h u am, c6 2 nguon
Ln=101og
am, giong nhau v a i cong suat phat am khong doi. Tai d i e m A c6 m i i c cuong
do am 20dB. De tai t r u n g diem M cua doan O A c6 m i i c cuong do am la
30dB t h i so'nguon am giong cac nguon am tren can dat t h e m tai O bang
A . 4.
B. 3.
C.5.
D . 7.
L M - L A = 1 0 1 o g - ^ = 101og
.OM)
= 10 log
20M
[OM
I0
loj
= 6 0 - 5 0 = 10
;
= 1=> — = 1 0 = > n = 100
10
JO
Chpn dap an C
l o w . Cho rang c u t r u y e n tren khoang each 2 m , nang l u p n g a m b i g i a m
M u c c u o n g do a m tai M k h i tai O c6 hai n g u o n am:
OA
^L„-Lio=101og
'nl
Vi d u 18: C o n g suat a m thanh cue dai cua m p t m a y nghe nhac gia d i n h la
'Phdn tick vd hu&ng dan gidi
I
log
If
^101^
6% so v o i Ian d a u do s u hap t h u cua m o i t r u o n g t r u y e n a m .
= 10 log 4
Biet lo =
10-'2W/m2.
N e u m o to het m u c t h i m u c c u o n g dp a m 6 khoang
each 6 m la:
L M = L A +101og4 = 20 + 101og4 = 26dB
A . 102,6 dB
G o i P va I la cong suat va c u o n g do am ciia m o i n g u o n .
B. 1 0 7 d B
C. 99 dB
D . 88 d B
'Phdn tich vd hitong ddn gidi
P
P
K h i do ta c6: I = — =
S
4nR^
am nen nang luang dm gidm dan trong qud trtnh truyen. Degidi
Goi I M , L M la cuong do a m va muc cuong dg am tai M k h i tai O c6 2 nguon diein.
todn ndy chung ta phdi ndm dugc moi lien he giita nang luang dm vd cong suat
G o i I ' M , L ' M la c u o n g d o a m va m u c c u o n g do a m t a i M k h i t a i O c6 n
dm: E = Pt
nguon diem.
Theo bai ra: C u 2 m t h i nang l u p n g a m g i a m 6% so v o i nang l u p n g ban dau,
= l O l o g JM. = 30 - 26 = 4 =^ ^
L;^ -
Ta co:
T
I M =
2P
-—j
47rR
i ;M
T
Day la bdi loan md trong qud trinh truyen am c6 su hap thu dm. Vi co su hap thu
— = — = lo"'"*
.
va
k h i do ta co:
= 10°'''
Sau 2 m d a u tien:
nP
IM =
= 0,94
Sau 2 m tiep theo (4m): = > - ? - = (0,94)^
n = 2.10°'^=5
=> Tai O CO tat ca 5 n g u o n am nen suy ra can dat t h e m 5 - 2 = 3 n g u o n a m
de tai M c6 m u c c u o n g do a m la 30dB. Chgn B
Sau 2 m tiep theo (6m): =>
T r o n g m o t b u o i hoa nhac, k h i d u n g 10 chiec ken d o n g t h i tai cho ciia m o t
khan gia d o d u g c m u c c u o n g do am 50dB. H o i p h a i diang bao nhieu
chiec ken d o n g de tai cho k h a n gia do c6 m u c c u o n g d p a m la 60dB?
B. 80
C.lOO
= (0,94f
Cong suat a m tai v j t r i each n g u o n am 6 m la:
Vl du 17: (Trich de thi thu chuyen nguyen Trai - Hai D u o n g Ian 1 nam 2013)
A . 50
= 0,06 ^ ^
47iR'
2P " 2
ntra
dugc dang bdi
D . 90
Pnt
= ( 0 , 9 4 ) ^ = > P 3 = ( 0 , 9 4 f Po
P
Pn.(0,94)^
C u o n g d p a m phat d i t u n g u o n d i e m d u p e xae d i n h la: I3 = -3- = '2—
S
47td
Vay m u c c u o n g cto am tai v i t r i each n g u o n am 6 m la:
T u ( l ) v a (2)
L2 - L , = 1 0 1 g ^ - 1 0 1 g i L
= ioigIl
V i : U - L ] = 2 0 = ^ 2 0 = 1 0 1 g ^ ^ ^ = 102 =100.
C h o n dap an A
'1
' '
c a u 4 : ( T r i c h de t h i t h u c h u y e n n g u y e n T r a i - H a i D u o n g Ian 1 n a m
ta B A I T A P V A N D U N G
C a u 1 : ( T r i c h de t h i thtr c h u y e n dai hQC V i n h Ian 3 n a m 2012)
^
A. D p cao
gan lien v o i do thj dao d o n g v i the cac nhac cu khac nhau k h o n g the c6
D. CO am s^c p h u thuoc vao dang d o thj dao d o n g cua am.
Cling do t h i dao d o n g v i the k h o n g the c i m g am sac d u g c nen chi co dap an
'Phdn tick vd hic&ng ddn gidi
C dung.
A m do m o t chie'c dan bau phat ra
C h p n dap an C
A . (sai) v i am nghe cang t r a m k h i bien d o am cang nho va tan so am cang
C a u 5 : ( T r i c h de t h i t h u T H P T N h u T h a n h - T h a n h Hoa Ian 2 n a m
le« nho.
2013)
M u c c u o n g d o am L cua mot am c6 cuong do am 1 d u g c xac d i n h being cong
B. (sai) v i am nghe cang cao k h i tan so am cang Ion k h o n g phai m u c cuong
thuc (In la c u o n g do am chuan)
dp a m cang Ion.
A. L(dB) = 1 0 . l g ^ .
C. (sai) v i am ccS do cao p h u thuoc vao tan so.
D. ( d i i n g ) v i am ScIc la m o t dac t r u n g sinh ly ciia am va p h u thuoc vao dang
do t h i dao d o n g ciia am.
C h p n dap an D
C a u 2 : C u o n g do am chuan la IQ = 1 0 " ' ^ W / m ^ C u o n g d o am tai m o t diem
trong m o i t r u o n g t r u y e n am la l O ^ ^ W / m ^ . M u c c u o n g am tai d i e m d o bang
B.60dB
C.70dB
D.80dB
'Phdn tich vd huang ddn gidi
= 10 Ig
in^^
10
= 10 Ig 10^ = 10.7 Ig 10 = 70dB. C h p n dap an C
10 ^ I
I
C. L(dB) = 10.ig
D. L(dB) = — I g i - .
10
Phdn tich vd huang ddn gidi
C. 100
C h p n dap an A
C a u 6 : ( T r i c h de t h i t h u c h u y e n D a i Hpc V i n h Ian 1 n a m
v a i l 2 t h i L2 = 1 0 1 g ^
(1)
(2)
being
nguon d a n g h u o n g va m o i t r u o n g k h o n g hap thu am. K h o a n g each t u
nguon am den d i e m ma tai do m u c c u o n g do am b5ng 0 la
B. 250m.
C. 500m.
D . 1000m.
Phdn tich vd huang ddn gidi
*•
Ggi B la d i e m c6 m u c c u o n g do am bang 0
Theo bai ra ta c6:
v 6 i l i thi L j = l O I g ^
2013)
10 *'W/m'^. C u o n g do am chuan b^ng 10 '-^VV/m^. Cho rang n g u o n am la
D . 1000
<Phdn tich v>d huang ddn gidi
ho
M u c c u o n g d o am d u g c xac d j n h theo cong thuc: L(dB) = l O . l g y
A. 750m.
am cua c h i i n g la
B. 20
—uA.
B. L(dB) =
C u o n g d o am tai d i e m A each mot nguon am d i e m mot khoang I m
C a u 3 : Hal am c6 m u c c u o n g do am chenh lech nhau 20dB. T y so c u o n g do
T a c 6 : L = 101g—
D. Ca A , B va C
an CO dap an D la dap an d u n g cho ca 3 dap an con lai. Chu y rSng: am ScIc
C. CO do cao p h u thuoc vao h i n h dang va kich thuoc hop cong h u o n g .
A . 10
C. A m s5c
Cau nay tha'y ngan the nay n h u n g hau het cac ban deu boi roi v i t r o n g dap
B. nghe cang cao k h i i p u c c u o n g d o am cang I o n .
t
Ta ccS: L = 10 Ig —
1()
B. D o to
'Phdn tich vd hii&ng ddn gidi
A. nghe cang traim k h i bien do am cang nho va tan so am cang Ion.
'
2013)
A m do hai nhac cu khac nhau phat ra luon khac nhau ve:
.
A m do m o t chie'c dan bau phat ra
A.SOdB
C h p n dap an C
M
LA-LB=201og
C h p n dap an D
:
R
= 60 => log
A
J
= 3 = ^ R B = 10-^RA = 1000m
C a u 7 : (Trich de thi t h u T H P T T h i ? u H o a - T h a n h H o a Ian 2 n a m 2013)
Phat bieu nao sau day khong dung?
= ^a -
I - 02 01
:BJ
iBq
IOA
d6\\^
I'f'
A . D o cao ciia a m la m o t dac ti'nh sinh If ciia a m .
B. Nhac am la d o n h i e u nhac cu phat ra.
C. Tap a m la cac a m c6 tan so'khong xac d i n h .
f V
OZ
^Ho2 0 l =
Đo[<=i't7 = 6'se-ot' =
ã
D . A m sac la m ^ t dac t i n h sinh l i ciia am.
•
3
'Phdn tick vd huang ddn gidi
a
»
e
—•
O
V
A . (dung) v i d o cao cua a m la m o t dac t i n h sinh l i ciia a m gan lien v o i tan so
^£9 a
am.
B. (sai) V I nhac a m la d o m p t hay nhieu nhac cu phat ra d e u
D . (dung) vi am sic la m p t dac t i n h sinh li cua a m gan lien v o i d o thj dao
UI8ZV
?I D9 ^•0!^ M^?^ 8uBoq>f
dugc.
C. (diing) v i theo d i n h nghla: Tap a m la cac am c6 tan so k h o n g xac d i n h .
uigorg
luOt'-D
BriiS MDBD 8uBoij>i gpoe
onyi
gp6'ge
BA
UIB uonSu UBp B X
'.qpof
BJ
]6ni
UB[ J
'g ' y
B A UIQ£
IB; UIB
op
B[
gy
Suqno
^ q ; oaq; 'Sueq §ugq; j 'g ' y uiaip e o q ^ -.jj
nej
a U B dep u 6 q 3
d o n g a m.
aPOOl = 03 + 08 = 03 + "^1 = ^1 <
=
ChQn dap an B
/
C a u 8 : Ba d i e m O , A , B cung n a m tren m o t nua d u o n g t h a n g xua't phat tu O.
03 =
Tai O dat m o t n g u o n d i e m phat song a m dang h u o n g ra k h o n g gian, m o i
.01
§0103 =
01
I
Soi03 =
a
a
go[oi =
8oio[= o i l -
'I
t r u o n g k h o n g hap t h u am. M i i c c u o n g d p a m tai A la 70 dB, tai B la 30 dB.
:oD B ; BJ iBq oaqj^
M i i c c u o n g d o a m tai t r u n g d i e m M ciia doan A B la
A.45dB.
B.54dB.
C. 36 dB.
D . 75 dB.
apool a
^hdn tick v>d huomg dan gidi
SuoriD Djiui i q i u i i
^
^' = 10^ =
• Rg = I O O R A
IB
AO
T ,
M la t r u n g d i e m A B ^ R M =
RA+RB
=
RA+100RA
^
IM
LM=LA-101og
noi
RM
R A ;
= 70-10.1og
= 10. log
ion
^
ion
UIB
uonSu qDBD uiaip
uonSu qDBD u i a i p
40UI
IBX
"J?
IBX
•gpo8
ex UBqd
BA
ej
UIB
o p S u o n j oxxui i q ;
riqi d s q x\s
00
8uoq>i ris
B I 3 uBig Suoq>f guoj4 S u o n q § u B p U I B 4Bqd u i a i p uonSu G\B uon§u ]6y^
i q ; ap q D u ^ ) : 0 T
nej
^666^
§0101 =
^ O£ 6 6 6 ^
^" r ^ ^ ^ A
^1
OT
8oroi = -rj§oroi = 9 . i
: A B U Dn[ g I B } U I B o p g u o r i D D J I J / ^
666 X'neee'^i'
^QV^P ^
-v— = —
== ^I
d
d
'
:v\u Drq g IB4 U I B 6p SuoriD i q i y 1B4 4 B p uie uonSu i\\y{
= 36dB.
truyen a m c6 m i i c c uo n g dp a m tai A va B Ian l u g t la: L A = 80 dB;
N e u n g u o n a m d o dat tai A t h i m u c c uo n g do am tai B k h i d o la
C50
UIB
apos = o8+oe-= "^1+
C a u 9 : H a i d i e m A v a B n a m 6 cung 1 phia ciia n g u o n am, tren c i i n g 1 phuonj^
B30
aP06 V
3 U B dep u 6 q 3
C h g n dap an C
A 20
"JOI
(£103 u i ? u £ u ? i q u j N D ? g u3Xnq3
IOIRA
- = —
H i e u m i i c c u o n g d o a m tai A va M
LA-LM=10.1og^-10.1og
gpoiig
SuBq UIB o p
Theo bai ra, ta c6:
L A - L B = 1 0 . 1 o g i ^ = 70 - 30 = 40
apo3iD
D 40
va666= ^ a - 3 a = a v ^ ^ a o o o i = 3 a
LB = 20 ciP
a' I
09 = 03 - 08 = v j S o i o i = ^ 1 - ^ 1
iviS uyp Sujpnif ya
uyii^
9
a
= oOl =
B4 'BJ iBq o a q i
'Phdn tich vd huong ddn gidi
^ ,j
Gpi P va I la cong sua't va cuang dp am aia moi nguon.
P
P
Khi do ta c6: I = — =
r
S 4nR^
Gpi IM ,LM la cuong dp am va muc cuong dp am tai M khi tai O c61 nguon die'm
Gpi I'M ,L'M la cuong dp am va muc cuang dpamtaiMkhitaiOcon nguon diem
L M - L M = 1 0 . 1 o g ^ = 6 0 - 4 0 = 2 0 = ^ ^ = 20
nPi
= n = 20
>-v .
Tgi O CO tat ca 20 nguon am. Chpn dap an A
Cau 15: Mpt nguoi dung giua hai loa A va B. Khi loa A bat thi nguoi do
nghe dupe am c6 muc cuong dp 60dB. Khi loa B bat thi nghe dupe am c6
muc cuong dp 70 dB. Neu bat ca hai loa thi nghe dupe am c6 muc cuang dp
bao nhieu?
A. 130dB
B. 70,4dB
C. 60dB
D. 70dB
'Phdn tich vd huang ddn gidi
Gpi Ij va I2 la cuang dp am loa A va loa B tai diem do. Khi do cuang dp
am tai diem do khi bac ca hai loa la: I = Ij +12
Theo bai ra, ta c6:
Muc cuang dp am cua cua loa A: Lj = lO.log— = 60 => Ii = IO^IQ
Muc cuong dp am cua loa B:
= lO.log 1?- = 70 => Ij = 10^IQ
Miic cuong dp am khi bac ca loa A va loa B:
L,2
= 1 0 . 1 o g i l ^ . l 0 . 1 a g l ^ i l I ^ = 70,4dB
Chpn dap an B
Cau 16: Hai hpa am lien tiep do mpt day dan phat ra c6 tan so han kem nhau
la 56Hz. Hpa am thu 3 c6 tan so la
A.28HZ
B.56HZ
C. 84Hz
D. 168Hz
i'hdn tich vd hucmg ddn gidi
Theo de ra, ta c6 nf - (n -1)f = 56 => tan so am ca ban f = 56Hz
Tan so hpa am thu 3: £3 = 3f = 3.56 = 168Hz. Chpn dap an D
= 4 0 - 3 0 = 1 0 => log
LA-Lc=201og
1
' W
1
Rc
1.
RA
2
••• > •
' „•!
I:U
-•• T trjPt:
•
M ,
;
=>Rc= 102 RA = 102.50 = 158m
Vgy BC = R c - R B = 1 5 8 - 8 0 = 78m
Chpn dap an A
Cau 12: Tai mpt diem nghe dupe dong thai hai am: am truyen toi c6 muc
cuang dp am la lOdB, am phan xa eo muc cuong dp am la 20dB. Muc cuong
dp am toan phan tai diem do la?
A.50,5dB
B. 12,5dB
C. 46,9dB
D. 20,4dB
^hdn tich vd hudng ddn gidi
Gpi Ij va I2 la cuang dp am tai va cuang dp am phan x^ tai diem do. Khi
do cuong dp am toan phan la: I = Ij +12
M
Theo bai ra, taco :
Muc cuang dp am ciia am truyen toi: Lj = lO.log— = 10 => Ij = IOIQ
Muc cuong dp am cua am phan xa: L2 = lO.log — = 20 => I2 = IOOIQ
Muc cuong dp am toan phan:
L,2 = 1 0 . 1 o g l i l i l = 10.1ogl^^5^ti^==20,4dB
Chpn dap an D
CSu 13: Trong mpt ban hpp ca, coi mpi ca si deu hat vai cung cuong dp am
va cung tan so'. Khi mpt ca si hat thi muc cuang dp am la 60dB. Khi ca ban
hpp ca cung hat thi do dupe muc cuong dp am la 90dB. So ca si c6 trong
ban hpp ca la
D. 1000 nguoi
A. 10000 nguoi.
B. 100 nguoi.
C. 10 nguoi.
Phdn tich vd hit&ng ddn gidi
Gpi so' ea si la n, cuang dp am cua moi ca si la I
Suy ra cuong dp am eiia n ea si se la: = nl
L„ - Li - 10.1og ^ = 10.1og^ = lO.log n = 90 - 60 = 30: . n = 1000
I1
Chpn dap an D
11
A
Cau 14: Tai O eo mpt nguon am nho phat song am den M thi tai M ta do
dupe muc cuong dp am la 40 dB. Neu tai M do dupe m u c cuang dp am la
60 dB thi tai O ta phai dat tong so nguon am giong nhau la
A. n = 20 nguon
B. n = 50 nguon
C. n = 10 nguon
D. n = 100 nguon
S
Cty TNHH MTV DWH
Cau
Khang Viet
19: Song (A, B cung phia so v o i S va A B = 100m). D i e m M la t r u n g dieVn
„uye't on luyftt thi dai hqc dat diem toi da Vat U, tap 1 - Le Van Vinh
Cau 17: M o t
6'ng k h i c6 m p t d a u bjt k i n , mpt dau h o tao ra am co b a n c6 tan
A B va each S 70 m c6 m u c c u o n g dp am 40dB. Biet van to'c am t r o n g khong
so n 2 H z . Biet toe d p t r u y e n am t r o n g k h o n g k h i la 336m/s. Buoc song dai
k h i la 340m/s cho r^ng m o i t r u o n g k h o n g hap thu am (cuong d p am chuaVi
nha't cua cac hpa am ma o n g nay tao ra b i n g :
Ii, = lO'^W/m^). N a n g l u p n g ciia song am trong k h o n g gian g i o i han b o i hai
mat cau tarn S qua A va B la
A . 207,9nJ
B. 207,9 m j
C. 2 a 7 m J
B. 0,8 m .
C. 0,2 m
D i e u kien de CO song d u n g t r o n g ong:
Song t r u y e n t r o n g k h o n g gian. N a n g l u p n g song t i I f nghjch
muc
c u o n g d p am tai d i e m M la t r u n g d i e m AB, nghla la se xac d j n h d u p e cuong
d p am tai M . Can c u suy ra c u a n g dp am tai A va B. C u o n g d p am tai A va
B t i le nghjch v o i b i n h p h u o n g khoang each d o n v j la W/m ^
X
1 = (2k + 1)—=>
4
voi binh
d a y de y cho
D. 2m.
(phdn tich vd himng dan gidi
D . 2,07^J
'Phdn tich v>d huong dan gidi
p h u o n g khoang each. N a n g l u p n g song b i n g gi? 6
A. I m .
Nang lupng
\
41
=
*
(*)
'
2k + 1
(I la chicu dai cua cot khi trong 6'ng, dau kin la nut dau ha la bung cua song dinig
trong ong khi)
. f = = ^ = (2k + l ) ^ = (2k + l ) ^
(f,, = — : tan so am co ban)
"
41
song tai cac m a t cau tam (S, SA) va (S, SB). Lay h i f u t h i d u p e nang l u p n g
Bai ra ta co: t , = 112Hz => ^ = 112
41
t r o n g v u n g g i o i han.
• = 0,75m
4.112
A m CO ban u n g v o i k = 0.
Theo gia thiet:
•"A = I'M -
AB
T u (*) ta tha'y cac hoa am co X^^^ k h i {2k+
2
AB,
Vay:
C u o n g d o a m tai 1 d i e m la nang l u p n g d i qua m p t d a n v j d i f n tich tinh
P=lM.47ir,
N a n g l u p n g t r o n g h i n h cau tam (5, SA) va (S, SB) la:
Cau
l,r,^^=j
C h p n dap an A
C h p n dap an A .
= 1{m).
18: Tai m o t d i e m tren mat phang chat long co m p t n g u o n dao d p n g tao
ra scSng o n d i n h tren mat chat long. Coi m o i t r u o n g tuyet d o i d a n h o i . M va
N la 2 d i e m t r e n m|it chat long, each n g u o n Ian l u p t la R i va R 2 . Biet bien d p
dao d p n g cua phan t u tai M gap 9 Ian tai N . T i so
B. 1/16
. A . 81
• W = W3 - W ,
1)^-^ = 3 ( v o i k = 1)
41
trong 1 d o n v j t h o i gian. T u gia thiet suy ra cong suat n g u o n S la
V „
1=
^^ang
C. 1/2
D. l/81i
'Phdn tich va hu&ng dan gidi:
v
10-^.471.75^
= l M i ! ^ ( r , - r , ) = ^ ^ - ^ ( 1 0 0 ) . 207,9,J
340
N a n g l u p n g song co t i le v o i b i n h p h u o n g bien d p , tai m p t d i e m tren mat
phang chat long co m p t nguon dao d p n g tao ra song on d j n h tren mat chat
l o n g thi nang l u p n g song truyen d i se dupe phan bo deu cho d u d n g tron
(tam tai n g u o n scSng). C o n g suat t u nguon t r u y e n den cho 1 d o n vj dai v o n g
tron tam O ban k i n h R la
27iR
2TrR
Suy ra
EN
RM
R.
_JQ__
RM
R,
27IRN
Vay ^
= AM =9^=81
C h g n dap an D
R,
81
Cty TNHH MTV DWH Khang Vift
.
,
. . ,
ZL-ZC
200-100
,
7
1
.
Do lech pha e i u a u va i: \ai\(D = ——^=
=l=::>(p = -rad
100
4
Tu do ta de dang tinh dugc pha ban dau cua hi^u di#n the hai dau mgch:
7t
Si (juye't on luy^n thi dai HQC dat diem toi da Vat It, tap 1 - Le Van Vinh
(phdn 3:
Chnyto
D d N G CII$N X D A Y C H I ^ U
1
B A I T O A N M± V I B I ^ U
71
cPu =cpi+(p = 0 + - = -
: •
THCrc
'Phucm.g phdp:
Bieu thuc u hoac i se luon c6 dang:
Bieu thuc hi^u di^n thehai dau mach: + 1007tt
(V)
4J
u = Uo cos((ot + (Pu) = 2Q0\[l cos
Bieu thuc hieu di?n the hai dau dif n tro R:
u = Uo.cos(cot + cpy) hoac: i = Io.cos(cot + cpi)
Vi vay de viet dupe bieu thuc cua hieu dien the hoac cuong dp dong di$n
chiing ta can phai xac dinh 4 yeu to' la Uo, lo, co va cp. Sau do diing cong
thuc: cp = cpu - cpi
U R = UoRCOs(cot + (PuR ) ; Voi: UOR= IO.R = 2.100 = 200 V ;
Trong dean mach chi chua R thi u ^ cung pha i :
Vay: U R = UORCOS (cot + cPu^ ) = 200cos lOGTit V
Voi cp la dp l^ch pha giua hi^u dien the so voi cuong dp dong di^n.
Bieu thuc hi^u di^n the hai dau cupn cam L:
cp dupe xac dinh theo cong thuc: tgcp = — — ^ .
R
ii
Lttu y: Mach khuyet phan tu gi thi trong cong thuc tren, ta khong dua vao.
U L = UoLCOs(cot + cpuj^) V a i : UOL= IO.ZL = 2.200 = 400 V;
Trong doan m^ch chi chua L thi U L nhanh pha hon cuong dp dong di?n ^
71
-
7t
7t
Cdc trudng hop dac biet:
,
f
^
+ Doan mach chi chua di^n tro thuan R thi cp = 0
Vay:
+ Doan mach chi chua cupn thuan thi cp = +—
2
= 400cos 100Kt + - V
V
2^
Bieu thuc hi^u di^n the hai dau ty di^n C:
U c = Uoccos(cot + (Pu^) V o i : Uoc= lo.Zc = 2.100 = 200V;
UL
=UoLCOs(cot + (Pu^^)
+ Doan mach chi chua ty difn thi (p = - ^
Trong doan mach chi chua C thi xx^ cham pha hon cuong dp dong difn ^
71
m VI DU
MAU
v i du 1: Mot mach di^n xoay chieu RLC khong phan nhanh co:
7t
^ u c = < P i - f = 0 - ^ = - ^ rad
R = 100Q; C = - . 1 0 ~ * F ; L = - H .
7C
Vay: uc= Uoccos(cot + (p^,„) = 200cosflOOTtt--1V
2
Vi dy 2 (Trich de thi dai hpc 2013):
Dat di^n ap u =
22o72 cosl007it
(V) vao hai dau doan mach mkc nol tiep
10"^
gom di^n tro R = 100Q, tu dien c6 C =
F va cupn cam thuan c6
27t
71
Cuong dp dong dien qua mach co dang: i = 2cosl00nt (A). Viet bieu thuc
tuc thoi di^n ap cua hai dau mach va hai dau moi phan t u mach di^n.
^hdn tick m huang dan gidi
Day la bdi todn viel bieu thiic mang tinh tong quat, qua bai todn nay cdc ban se co
duoc phuang phdp deldm cdc bdi todn vesau cua dang nay.
Cam khang:
= L.co = -IOO71 = 200Q
7t
L = — H . Bieu thuc cuong dp dong dif n trong doan mach la
Dung khang: Zp = — =
A. i=2,2^/2cos 1007tt + C. i = 2,2 cos 1007lt + 4
(A)
(A)
B. i = 2,2cos l O O T l t - -
^ - — r = 100 Q
Tong tro: Z =
= ^100^+(200-100)^ ^ 100720
(A)
D. i = 2,2^2 cos lOOTtt-4j
(A)
+{Z^-Z^f
Hi?u di?n the cue dai 6 hai dau mach: Uo = lo.Z = 2. IOOV2 V =200 72 V
