PHÂN LOẠI BÀI TẬP CHƯƠNG DAO ĐỘNG CƠ
Dạng 1: Vận dụng các đặc điểm của dao động điều hòa, so sánh pha của dao động.
Câu 1: !"#$!
%&
ω
=
'()*!+,-./&0'(1!+,234/
5
0'()6
3!7 ,)8)/&
5
09)%:0);0<)=0
Câu 2: !"#)+,>!?6@34-.5%)=0'(.!+AB!,-.=0'(
/
)
CDπ
/
≈
%&)6E(+2.3!7 ,)
Câu 3: Một !" .7?FB 2!".!=&G0H)*!I6JK%&G0H2+
H'G//& scmv
π
=
)
>L ,-.M 8)%G(H9)&NOG(H)&N%G(H<)OG(H
Câu 4: !,0P+ !" . J!-7 B89K%:G0H.-.0 Q>R,
P?$!+%/&&G#'SH)9T?0U!0(V)>L .+AB!,S!-.M
8)
H'G/N5HWG
/&
%
sms
π
9)
H'G/N:5HWG/& sms
π
)
H'G5/HWG
/&
%
sms
π
<)
H'G5/HWG/& sms
π
Câu 5: !"#$!E(+2
/&
ω
=
'(N !" .$!3!78K:0)U+X!!-
!?6@34)YZX!Z[
%&
π
(E!7-.M8):0) 9)/=0))\0) <)%/0)
Câu 7: D!]0 !"#7?FB 2!".!/&0. > X!!5S2A!^O=&
.SE)63!7.E(+ )A)%&0W5_B)/&0W%_C)%&0W/_D)/&0W5_
Câu 8M !"#$!SZP1MJK:(!Gπ`HG0H)C!.+,IX!!]0K(-.M8)JK
:0WK&9)JK50WK5π0'()JK50WK5π0'(<)JK50WK5π0'(
Câu 8M !"#$!3!7O0N>!2-!JKa5012+=π0'()E(+ -.M
8)O_ 9)/_)&N/_ <)&NO_
Câu 9: !"#2SZP1
= (G%& H
:
x c t cm
π
π
= +
). X!!]0K&I@.!]
b !". N+-.3 !7c
8)JK/0N
/& 5 'v cm s
π
= −
Nb !"@0)9)JK/0N
/& 5 'v cm s
π
=
Nb !"ZP)
)
/ 5x cm= −
N
/& 'v cm s
π
=
Nb !"ZP)<)
/ 5x cm=
N
/& 'v cm s
π
=
Nb !"ZP)
Câu 10MD!]02SJK8 (G
ω
H70ZXdK/&0N2>L K/()!e3!]f
+N!+.6V!VAB!,)
Câu 11Mg,0 !" .2BZ1hM
9!7N.S3E-E-Z[-.M
8)=0W&)
9)a=0Wai)
)=0Wi)
<)a=0W&
Câu 12Mg3!]!(A3!e!7,+b -! !" .21B. (@M
A. ZXS3 -WB. ZX#WC. ZXb-!SWD. ZXSb3 -
Câu 13. !" .>!2-!
%
/x cm=
1+
%
= 5v
π
=
0N>!2-!
/
/ /x cm=
12+
/
= /v
π
=
0)9!7.E(+ ,-.M
A.=0.%_) B.;0./_) C.
= /cm
./_) D.VSV>V)
Câu 14. !" . j>L!Z[?YZX%&0)*!2-!JK5012+
K%:π0'()>L ,-.MA.&NO(B.%N:(C.%(D./(
Câu 15: !" .$!SZP1JK= (G=π`π'5H)6?YZX-$D0.!Z[
> X!!∆K%':G(HMA.=
5
0B.5
5
0C.
5
0D./
5
0
@%O)/M !"#Ub QkJN?6@34k$!3!78.>L) >
X!!'5N?YZXTD0.2]!Z[-.M8)G√5a%H89)8)8)√5<)8)G/a√/H
Câu 16: !" .N>!2-!J
%
K=01+
%
=& 5 'v cm s
π
= −
W>!2-!
/
= /x cm=
1+
/
=& / 'v cm s
π
=
)6>L MA.%):(B.&N/(C.&N;(D. &N=(
Câu 17: !" .$!SZP1-!JK%&(!G;πaπ'5H0)*!?62-!:01
+,2-.M8):=π0'(9)±;&π0'()±:=π0'( <);&π0'(
Câu 18: !" .
8)+3!el!!" .mS( $!-!) 9)+3!el!!" .Z[S( $!-!)
%
)+3!el!!" .($0Sπ'/( $!-!)<)+3!el!!" .0Sπ'/( $!-!)
Câu 19: !" .
8)!+3!el!!" .mS( $!-!) 9)!+3!el!!" .Z[S( $!-!)
)!+3!el!!" .($0Sπ'/( $!-!)<)!+3!el!!" .0Sπ'/( $!-!)
Câu 20: !" .
8)!+3!el!!" .mS( $!+)9)!+3!el!!" .Z[S( $!+)
)!+3!el!!" .($0Sπ'/( $!+)<)!+3!el!!" .0Sπ'/( $!+)
Câu 20:D!]0 !"#)*!!?6@34N+,D!]0-.=&0'(NB!63!7!
+2-$/&&0'(
/
)9!7 ,D!]0-.MA. &N%0)B. ;0)C. O0)D. &N;0)
Dạng 2: Viết phương trình của dao động điều hòa
Bài 1M !"#$!>LK%(.3!78K%&0)!eSZP1 , VZX
[S(M
HU+X!!K&-2-JK8G63!7ZPH
3HU+X!!K&-2-JKa8G63!7@0H
HU+X!!K&-!?6@34Mb !"ZP.!"@0
HU+X!!K&-2-JK
/
A
)b !"ZP.!"@0
bHU+X!!K&-2-JK
/
A
−
)b !"ZP.!"@0
nHU+X!!K&-2-JK
±
/
/
A
)b !"ZP.!"@0
HU+X!!K&-2-JK
±
5
/
A
)b !"ZP.!"@0
HY10?-,!^!eSZP1 .3!]!27QU)
Câu 2. !"#$!
O
ω
=
'()B!6@34" 0+%NO0'(b !"ZP)
ZP1 -.M
8)JK&N5(!GO`π'/H09) JK&N5(!GOH0 C. JK&N%O(!GOaπ'/H0<)JK&N%O(!GOH0
Câu 3. !"#$!
%& /
ω
=
'() +X!!K&-2-JK/
5
0.!"
6@34$!+&N/
/
0'()CDK%&0'(
/)
ZP1 ,?E2BM
8)JK=(!G%&
/
`π'=H9)JK=(!G%&
/
`/π'5H)JK=(!G%&
/
`Oπ':H<)JK=(!G%&
/
`π'5H
Câu 4: $!3!7:G0H)CK&N -o?62-!JK5
/
G0Hb !"ZP$!!+2
-$
5
/
G0'(
/
H)ZP1 , -o-.M
8)JK: (\G0H9p)
J : (
5 =
π
= −
÷
G0H)
J : (
5 =
π
= +
÷
G0H<)
J : ( 5
5
π
= +
÷
G0H
Câu 5: D!]02>+!-Z[0K%& !"#7 Bd.!=0NE(+O_)CK&ND!]0I
6 @ 34 . 3o E ! b Z$ ZP , ?F B ) 9!] f U , b X! !M
8)JK/ (G%&iai'/H0 9)JK/ (%&i0 )JK= (G%&i`i'/H0 <)JK= (Oi0
Câu 6: 2>+!-Z[0K%> !" .$!>1K/()?6@34$!+
&
K5%N=0'()
*!K&N?62-!JKO0Z[!"ZP?B )CDπ
/
K%&)ZP1 ,-.M
Câu 7M ! !" .mSZPNm>1K/()< fD2-!IX!!]03EGK&H34
3!7 .34%0)< f!23!734
5
0NIX!!]03E-!34&.+2!V
@0)
!eSZP1 ,! Y )
8HJ
%
K/ (πG0HNJ
/
K
5
(!πG0H9HJ
%
K (πG0HNJ
/
Ka
5
(!πG0H
HJ
%
Ka/ (πG0HNJ
/
K
5
(!πG0H<HJ
%
K/ (πG0HNJ
/
K/
5
(!πG0H
Câu 8: -o-#J g0?ET.2f>K;&'0) -oA!^%&& e5%N=()U+X!
!-.-?E2-!/0.]b !"ZP,QU$!+2-$
=& 50 ' (
1
SZP1 ,?E-.M
/
A.
J = (G/&a '5H0
= π
B.
J : (G/&` ':H0
= π
C.
J = (G/&` ':H0
= π
D.
J : (G/&a '5H0
= π
Câu 9: 2>+!-Z[0K%> !" .$!>1K/()?6@34$!+
&
K
5%N=0'()*!K&N?62-!JKO0Z[!"ZP?B )CDπ
/
K%&)ZP1
!" .,-.M
Câu 15. !" .f(%';(1q-B!34eq)YZX!Z[ &NO(-.%:0)
U+X!!-?6@34b !"@0)ZP1 ,-.M
A.
; (G/ H
/
x c cm
π
π
= +
B.
; (G/ H
/
x c cm
π
π
= −
C.
= (G= H
/
x c cm
π
π
= −
D.
= (G= H
/
x c cm
π
π
= +
@%&M-#J rb df2!".!A!7-.5&0)b . EZ$!-#J 0T1D^@34
>!-#J !Y%&0)*s b SZPdf $!>!-#J 2!".!=/0Ng!" +/&0'(
Z$-77G !" .H)U+X!!>!Z["+N!"ZPZ$-7)CD
/
'%& smg =
)ZP1 ,-.M
8)JK
t%& (//
G0H9)JK
t%& (/
G0H)JK
H
=
5
%& (G//
π
−t
G0H<)JK
H
=
%& (G/
π
+t
G0H
Câu 46: !" .>!?6@342+K/&0'(.!+AB!,-.K/0'(
/
)
UK&-.-?6@34b !"@0,Q BNSZP1 ,-.M
A. JK/ (G%&H0)B. JK/ (G%&`πH0)C. JK/ (G%&aπ'/H0)D. JK/ (G%&`π'/H0)
Câu 37: !" .>!?6@342+K/&0'()t!+AB!,-.
0J
K/0'(
/
)
UK&-.-?6@34b !"@0,Q B)ZP1 ,-.M
A. JK/ (G%&`iH0)B. JK/ (G%&`i'/H0)C. JK/ (G%&i'/H0)D. JK/ (G%&H0)
/)ZP1 . 3!ef$!X!!]0K%NO(2-!JKaOG0Hc
8)JKO(!G5π`πHG0HW9)JKO(!/πG0HW)JKO(!G5π`π'/HG0HW<)JKO(!5πG0H
6.D!]0A!^ !" .b SZP407 Bd89K/$!>1K/()U+
X!!-K&N>!D!]040I-!JK'/.+2!V@0)ZP1 ,D!]02BM8)J
K(!Gπ`Oπ':HW9)JK/(!Gπ`π':HW)JK/(!Gπ`Oπ':HW<)JK(!Gπ`π':H
Câu 17:2>+!-Z[0K%> !" .$!>LK/()?6@34$!+
&
K&N5%=
0'()*!K&?62-!JKO0b !"@0,?FB )CD
/
π
K%&)ZP1 !" .,
-.MA. JK%& (G
π
`
5
π
HB. JK%& (G=
π
`
:
π
HC. JK%& (G=
π
`
:
O
π
HD. JK%& (G
π
`
:
π
H
Câu 49: -o-#J !" .$!>LKO()9!e4B!X!!]0KO(?-o2-!JK
/
/
0.
+K
)'
O
/
scm
π
ZP1 , -o-#J 2BZe. c
A. JK
/
(
−
/O
/
ππ
t
B. JK
/
(
+
/O
/
ππ
t
C. JK (
−
=O
/
ππ
t
D. JK (
+
=O
/
ππ
t
001: !" .>!?6@342+K/&0'()t!+AB!,-.
0J
K
/0'(
/
)UK&-.-?6@34b !"@0,Q B)ZP1 ,-.A. JK
/ (G%&H)B. JK/ (G%&`i'/H)C. JK/ (G%&`iH)D. JK/ (G%&i'/H
005: !" .f(%';(1q-B!34eq)YZX!Z[ &NO(-.%:0)
U+X!!-?6@34b !"@0)ZP1 ,-.M
A.
; (G/ H
/
x c t cm
π
π
= +
W B.
; (G/ H
/
x t cm
π
π
= −
W C.
= (G= H
/
x c t cm
π
π
= −
W D.
= (G= H
/
x c t cm
π
π
= +
W
Dạng 3: TÍNH THỜI GIAN ĐỂ VẬT ĐI TỪ VỊ TRÍ CÓ LY ĐỘ X
1
ĐẾN X
2
Câu 1: !"#$!>L.3!78)Y6> X!!oD]!u62-
HJ
%
K8eJ
/
K8'/3HJ
%
K8'/eJ
/
K&HJ
%
K&eJ
/
Ka8'/HJ
%
Ka8'/eJ
/
Ka8
bHJ
%
K8eJ
/
K8
/
5
nHJ
%
K8eJ
/
K8
/
/
HJ
%
K8eJ
/
Ka8'/
Câu 2M !"#$!3!78K=02>L K&N%(
8)6> X!!oD]!u62-J
%
K/0eJ
/
K=0
9)6> X!!oD]!u6J
%
Ka/0eJ
/
K/0
)6> X!!oD]!u6@34e6JK/0
5
Câu 3MMU!%-.X!!oD!u9e-!JK8'/./-.X!!!u6-!JK8'/
e3!7ZP)2M8)
%
K&NO
/
9)
%
K
/
)
%
K/
/
<)
%
K=
/
Câu 4M !"#2SZP1JK; (%&i)vVX!!]0!?6JK=-Ef/b !"
@0>]uX!!]03oE )
Câu 5: 0 !"#2SZP1]
π
−π=
:
/%& (J
G0H)!?6@34-E
E!7. X!!]0M 8p)
5
%
G(H 9)
:
%
G(H )
5
/
G(H <)
%/
%
G(H
Câu 6: -o-#J $!3!78)X!!oD]!u6@34e!]02-!
/
/8
J
=
-.
&N/OG(H)>L, -oM8)%G(H 9)%NOG(H )&NOG(H <p)/G(H
Câu 7M !"#$!SZP1
%&(!G H
/ :
x t cm
π π
= +
X!!oDu-3oE e
-?62-!
O 5cm−
-Ef5b !"ZP-.M8)w() 9)\() )%%()<)%/()
Câu 8M !"#2SZP1JK; (%&i)vVX!!]0!?6JK=-Ef/&&\>]u
X!!]03oE )
Câu 9M !"#2SZP1JK; (%&i)vVX!!]0!?6JK=-Ef/&&;b
!"@0>]uX!!]03oE )
Câu 10M -o-#J !" .70xSd$!>1K%NO(.3!78K=0NS3E-.
:'Oπ
)
6u-K&N2 BJKa/0-Ef/&&O. X!!]0. M
8)%O&5( 9)%O&5N/O( )%O&/N/O( <)%O&5N5wO(
Câu 11: -o-#J b df)*66 -o !"#b SZPdf)>1.3!7
, -o-E-Z[-.&N=(.;0)UQJyJdf!"ZPZ$J+N+ BB!6@
34N+X!!K&>!?6@34b !"ZP)CD!+P!A K%&0'(
/
.i
/
K%&)X!!
oD>]u>!K&e>!-A.g!,-#J 2-$A!]-.MA. w'5&()B. 5'%&()C. ='%O()D. %'5&()
Câu 12: -o-#J dfg0x2>+!-Z[%&&.0-#J r2f>K%&&'0)*s
J+Z$!b SZPdfe6-#J Y=0g!" 20+
scm'=&
π
b SZPdf
uZ$!-7) ! !" .b SZPdf)X!!oD]]u6DSDe
6-#J 3s%NO0-.M8)&N/( 9)
s
%O
%
)
s
%&
%
<)
s
/&
%
Câu 13M -o-#J b SZP$!SZP1JK8 (Gω`ϕH)f(z> X!!34
.34π'=&G(H1q,34eq,-#J ) -o !" .$!E(+234M
8)/&)(
%
9);&)(
%
)=&)(
%
<)%&)(
%
Câu 14: D!]0 !"#$!>11q.eq,23!e!7.34(z
> X!!-.M8)/ 9) )'/ <)'=
Câu 15M -o-#J b df)*66 -o !"#b SZPdf)>1.3!7
, -o-E-Z[-.&N=(.;0)UQJyJdf!"ZPZ$J+N+UB!6@
34N+X!!K&>!?6@34b !"ZP)CD!+P!A K%&0'(
/
.i
/
K%&)X!!
oD>]u>!K&e>!-A.g!,-#J 2-$A!]-.MA.
/
5&
s
)B.
w
5&
s
)C.
%
5&
s
)D.
=
%O
s
)
Câu 16: !" .2E(+/_N3!7=0){0X!!]0. 2]b !"@0?
62-!/01(X!!]02%'%/(]b
A. !"@0?6@34) B. !"ZP?62-!a/0)
C. !"@0?62-!
/ 5cm
−
) D. !"@0?62-!a/0)
Câu 17: D!]0 !"#b SZP1JK5 (GOi`i':HGJ6340.634!@H)
0!@E!7uX!!]0K&ND!]0!?62-!JK`%0MA. w-E)B. :-E)C. =-E)D. O-E)
Câu 18M !" .$!-
HHG
:
O
ON& (G= cmtx
π
π
−=
2634G(H). X!!]0. (@
!?6JK/
5
0b !"ZP,Q BM8)K%G(H9)K/G(H)KO
5
%
G(H<)K
5
%
G(H
=
Câu 19: !"#$!3!]f-
H
5
ON& (G=
π
π
−=
tx
N 2Nx6340N634!@).
X!!]0. (@(h!?6
cmx 5/
=
b !"@0,QUM
A. ='5G(H B. OG(H C. /G(H D. %'5G(H
Câu 20:D!]0 !"#b SZP1M
J /NO ( %&
/
π
= π +
÷
G0H)10+31,
%>L M8)O&G0'(H 9)O&G0'(H)OG0'(H <)OG0'(H
Câu 21: 0 !"#2SZP1]
π
−π=
:
/%& (J
G0H)!?6@34-E
E!7. X!!]0M8)
5
%
G(H 9)
:
%
G(H )
5
/
G(H <)
%/
%
G(H
Câu 22: !"#$!>1N70 BdN!z!!]03!7.)U!"ZPue
N+UB!6@34kN0+X!!K&-.-!?!]0|, Bkb !"ZP)t!+,
34>-EfD. X!!]0M
A. K) B. K) C. K) D. K)
Câu 23: -o-#J $!3!78NX!!oD] -o!]u62-!J
%
Ka8e6
2-!J
/
K8'/-.%()>1 , -o-.MA. %'5G(H)B. 5G(H)C. /G(H)D. :G(H)
Câu 24: b SZP1JK/ (GOπ`π':H`%G0H) !@E!7>]u-3oE
!?62-!JK/0b !"ZPZ[0D-Ec
A. /-E B. =-E C. 5-E D. O-E
Câu 25: !"#$!SZP1JK8 (G`H)X!!oD>]u-3oE 2!
+340z!VAB!-.MA)tK9)K)K<)K
Câu 26M !"#u9e$!>1-.N6@34-.k)!]0,k9.kb fA-.
.)X!!]!b 0!"ue-.M8)'=9)'/)'5<)':
Câu 27) !" .$!SZP1JK/) (G/
π
a
π
'/H0)}X!!w':(>]uX!!]03E
!?6JK%0M8)/-E9)5-E)=-E<)O-E
Câu 28) -o-#J !" .$!SZP1JK8 (/
π
G0H)q.eq, -o34
-EE!7-.M8)%';(9)%'=()%'/(<)%(
Câu 29) -o-#J dfN>!b -#J !Y=0)*66 b SZPdf$!
3!7;01 0>1 X!!-#J 3s-.M8)'=9)'/)':<)'5
Câu 30. -o-#J b df2f%&'0N2>+!-Z[/ON-DK%&0'(
/
)9EZX!@
-7( -#J >3!eBg!r NU+X!!-3oE NQ Jdf
!"ZPZ$J+)q.eq,34. zX!!]0-.M
A.
5
;& =&
k
t
π π
= +
() B.
5
;& /&
k
t
π π
= +
() C.
;& =&
k
t
π π
= − +
() D. VS(+>V)
Câu 31: !"#2>1)eU+X!!K&-?6@34N1 j>1E
!7N+,34>IX!!]0M8)K';9)K'=)K':<)K'/)
Câu 32. -o-#J 2x$!>+!-Z[0K%&&.-#J 2f>K%&'0 $!3!7/
0) 0~!>1 NX!!0.xIV6@34-$P%0-.3 !7c
A.&N=%w( B.&N5%w( C.&N/%w( D.&NO%w(
Câu 33. !" .$!SZP1
%& (G ` '5H0x c
π π
=
)X!!6u-3oE GK&H
e>!!Z[?YZXO&0-.M A.w'5( B. /N=( C.='5( D.%NO(
Câu 34:D!]0 !" .2+34>B!!X!!]0-!7!eS-.
%
K/N/G(H.
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K/N\G(H)6u
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Dạng 4: Quãng đường vật đi được
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Dạng 5 : Tốc độ trung bình, tốc độ trung bình lớn nhất
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Dạng 6: Thời gian lò xo bị nén hoặc bị dãn
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Dạng 7: Tính chu kỳ. Cắt ghép lò xo
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Câu 8: Treo quả nặng m vào lò xo thứ nhất, thì con lắc tương ứng dao động với chu kì 0,24s. Nếu treo quả nặng đó vào lò xo
thứ 2 thì con lắc tương ứng dao động với chu kì 0,32s. Nếu mắc song song 2 lò xo rồi gắn quả nặng m thì con lắc tương ứng
dao động với chu kì: A. 0,192s B. 0,56s C. 0,4s D. 0,08s
Câu 9:90
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Dạng 8: Lực đàn hồi cực đại và cực tiểu, chiều dài cực đại cực tiểu
Câu 1: -o-#J f*b dfNE7+NEZ$!o)!Y,-#J B!6@34-.
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