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Chuyn 1 : Dao ng c
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CH1: DAO NG IU HA1. Dao ng iu ha+ Dao ng iu ha l dao ng trong li ca vt l mt hm csin (hay sin) ca thi gian. + Phng trnh dao ng: x = Acos(t + ).+ im P dao ng iu ha trn mt on thng lun c th c coi l hnh chiu ca mt im M chuyn ngtrn u trn ng trn c ng knh l on thng .
2. Cc i lng c trng ca dao ng iu ho: Trong phng trnh x = Acos(t + )th:Cc i lng ctrng ngha n v
A bin dao ng; xmax = A >0 m, cm, mm(t + ) pha ca dao ng ti thi im t (s) Rad; hay pha ban u ca dao ng, Rad; hay tn s gc ca dao ng iu ha rad/s.T Chu k T ca dao ng iu ha l khong thi gian thc
hin mt dao ng ton phn :T =2
=N
t
s ( giy)
f Tn s f ca dao ng iu ha l s dao ng ton phn thc
hin c trong mt giy .
1
f T
Hz ( Hc) hay 1/s
Lin h gia , T v f:=
T
2= 2f=>
2T ;f
2
Bin A v pha ban u ph thuc vo cch kch thch ban u lm cho h dao ng,Tn s gc (chu k T, tn s f) ch ph thuc vo cu to ca h dao ng.
3. Mi lin h gia li , vn tc v gia tc ca vt dao ng iu ho:i lng Biu thc So snh, lin hLy x = Acos(t + ): l nghim ca phng trnh:
x + 2x = 0l phng trnh ng lc hc cadao ng iu ha.xmax= A
Li ca vt dao ng iu ha bin thin iu
ha cng tn s nhng tr pha hn2
so vi vi
vn tc.Vn tc v = x' = - Asin(t + )
v= Acos(t + +2
)
-V tr bin (x = A), v = 0.-V tr cn bng (x = 0), |v| = vmax= A.
-Vn tc ca vt dao ng iu habin thin
iu ha cng tn s nhng sm pha hn2
so
vi vi li .- Khi vt i tvtr bin vvtr cn bng th vntc c ln tng dn, khi vt i tvtr cn bngvbin th vn tc c ln gim dn.
Gia tc a = v' = x = - 2Acos(t + )a= - 2x.Vc t gia tc ca vt dao ng iu ha lunhng v v tr cn bng, c ln t l vi ln ca li .- bin (x = A), gia tc c ln cc i:
amax= 2A.
- v tr cn bng (x = 0), gia tc bng 0.
-Gia tc ca vt dao ng iu ha bin thin iu
ha cng tn s nhng ngc pha vi li x(smpha
2
so vi vn tc v).
-Khi vt i t v tr cn bng n v tr bin, a
ngc chiu vi v ( vt chuyn ng chm dn)-Khi vt i t v tr bin n v tr cn bng,
a cng chiu vi v ( vt chuyn ng nhanhdn).
Lc ko v F = ma = - kx
Lc tc dng ln vt dao ng iuha :lunhng v v tr cn bng, gi l lc ko v (hiphc).Fmax= kA
- Chuyn ng nhanh dn : a.v>0, vF ;
- Chuyn ng chm dn a.v
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4.H thc c lp i vi thi gian:
+Gia ta v vn tc:2 2
2 2 2
x v1
A A
22
2
vx A
22
2
vA x
2 2v A x 2 2
v
A x
+Gia gia tc v vn tc:2 2
2 2 4 2
v a1
A A
Hay
2 22
2 4
v aA
22 2 2
2.
av A
2 4 2 2 2. .a A v
5.Ch :
51Cc h qu:
+ Qu o dao ng iu ha l 2A+ Thi gian ngn nht i t bin ny n bin kia lT
2
+ Thi gian ngn nht i t VTC ra VT bin hoc ngc li lT
4
+ Qung ng vt i c trong mt chu k l 4A.
5.2) Mt vi phng trnh cn lu :
sin( ) cos( ); cos( ) sin( );2 2
cos( ) cos( ); sin( ) sin( );
x Asin( t ) A cos( t ).2
x A cos( t ) A cos( t )
x A t A t x A t A t
x A t A t x A t A t
*Phng trnh c bit.
a)x a Acos(t +) vi a const
b)x a Acos2(t +) vi a const in :A
2 ; 2 ; 2.
5.3.Cch lp phng trnh dao ng :
2 2 22 2
2 4 2
t=0
t 0
2 t2 f;T ;
T N
v a vA x
xshift cos V 0 0
A
in : A
Ta VTCB : x a
Ta vtr bin : x a A
O-
+
+
x
a X
x
v
a
+
O A-A+
+
x
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cos= 0x
A(ly nghim "-" khi v0> 0; ly nghim "+" khi v0< 0) ;
(vi x0v v0l li v vn tc ti thi im ban u t = 0).
Cc bc lp phng trnh dao ng dao ng iu ho:* Tnh * Tnh A
* Tnh da vo iu kin u: lc t = t0(thng t0= 0) Acossin
xv A
Lu :+ Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0+ Trc khi tnh cn xc nh r thuc gc phn t thmy ca ng trn lng gic
(thng ly - < )*Phng php:
+Tm T: khoangthoigian tTsodaodong N
Tm f :sodaodong N
fkhoangthoigian t
- Cng thc lin h
1
2f T
Tn sgcmax max max
max
2
2
v a a
fT A A v
+Bin A:
22 2
2
vA x
; 2
2WA
k ; max max
2 2
v a chieudaiquydaoA
+Ch : 0 0
0 0
* 0 0 * 0 0
* 0; 0 ; * 0; 02 2
* ; 0 0; * ; 0
v v
x v x v
x A v x A v
6.Xc nh thi im vt i qua ly x0 -vn tc vt t gi trv0
6.1) Khi vt i qua ly x0th x0= Acos(t + )
cos(t + ) = 0x
A t= ? (Khi c iu kin ca vt th ta loi bt mt nghim t )
6.2) Khi vt t vn tc v0th v0= -Asin(t + ) sin(t + ) = 0v
A t= ?
63) Tm ly vt khi vn tc c gi trv1:2
2 2 1vA x
2
2 1vx A
6.4) Tm vn tc khi qua ly x1:2
2 2
1
vA x
2 2
1v A x
khi vt i theo chiu dng th v>0 v ngc li
7.Nng lng ca dao ng iu ho:
a) Thnng: Wt=1
2kx2=
1
2kA2cos2(t +)
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b) ng nng: W1
2mv2
1
2m2A2sin2(t +)
1
2kA2sin2(t +) ; vi k m2
c) C nng: 22 2 2 2 t1 1 1
W=W W m A kA m 2 f A2 2 2
= const
d) Ch : + Khi Wt W x A 2
2 khong thi gian Wt = W l : t
T
4
(Trong mt chu k c 4 ln ng nng v thnng ca vt bng nhau nn khong thi gian lin tip gia hai ln
ng nng v thnng bng nhau l4
T.)
+ Khi vt dao ng iu ha vi tn sf, tn sgc chu kT th Thnng v ng nng ca vt
bin thin tun hon vi cng tn sgc 2, tn sdao ng f =2f v chu k TT/2.
+Khi tnh nng lng phi i khi lng vkg, vn tc vm/s, ly vmt
+Ti vtr c W= n.W
tTa :
1
A
x n ; Vn tc :
1
n
v A n
+Ti vtr c Wt= n.WTa :1
nx A
n
; Vn tc :
1
Av
n
8. S 1: Tng qut v qung ng i, thi gianv nng lng:
cos
-A A2
0 A2
A 2
2
A 3
2 +A
T/4 T/12 T/6
T/8 T/8
T/12 T/24 T/24 T/129 S 2:quan h gia li v vn tc
maxv v
max
3v v2
max
2v v
2 maxvv2
v 0
x0 (VTCB) A
2 A 2
2 A 3
2
+A
W= 3 WtWmax = kA2
Wt= 0Wt= 3 WW= WtW= 0
Wtmax= kA2
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VNG TRN LNG GIC-GC QUAY V THI GIAN QUAYCc gc quay v thi gian quay c tnh t gc A
A
x
A 2A 22
A
32
A
3A
2
O 2
A 22
A 32
A
= + /2T/4
= +/4T/8
= + /6T/12
v = 0= 0
= + /3
T/6
= - /6
= + 2/3T/3
= - /2
= - /3
= - /4
= + 3/43T/8
= +5/65T/12
= -5/6
= - 3/4
= - 2/3
v=0=
T/2
V
V>
O
0
2
2kAW Wt=
Wd=
Wt=0
0
2
2kAW
3
4W 34W
3
4W
3
4W
1
2W
1
2 W 1
2W
1
2W 1
4W
1
4 W 14W
1
4W
2
2
kAW
Ly x: xAO A/2 2
3A
2
A
-A -A/22
A 3
2A
Vntc: 00 max2
v max 3
2
v max2
v max
3
2
v max2
v max
2
v
Gia tc: x- AOmax
3
2
a max
2
a 2A
max
2
a max 32
a max2
a max
2
a
S thi gian: x
T/4
T/8
T/4
AO A/2 2
3A
2
A -A -A/22
A 3
2A
T/6T/6
T/12T/12 T/12 T/12T/12 T/12T/24T/24
T/2
T/8
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CC DNG BI TP
Dng 1: Xc nh cc i lng trong dao ng iu ha.
A. Kin thc cn bn:
a phng trnh cho vdng osx c t . T , ,A ch :
sin os sin os2 2t c t hay t c t
B. Cc v d:
V d1: Xc nh bin , chu kv pha ban u ca cc dao ng iu ha sau:
) 4 os 2 ; ) 4 os 5 4sin 53
a x c t cm b x c t t cm
Hng dn :
2) 4 os 2 4 os 2 4 os 2 4 , 2 /3 3 3a x c t cm c t c t A cm rad s
2 2
1 ;3
T s
) 4 os 5 4sin 5 4 os 5 4 os 5 4 2 os 52 4
b x c t t cm c t c t cm c t cm
2
4 2 ; 5 / 0,4 ;4
A cm rad s T s
V d 2:Mt vt dao ng iu ha theo phng trnh: x = 4 )2/.2cos( t (cm)a, Xc nh bin , chu k, pha ban u ca dao ng.
b, Lp biu thc ca vn tc v gia tc.
c, Tnh vn tc v gia tc ti thi im t =1
6s v xc nh tnh cht chuyn ng.
Hng dn:a, A = 4cm; T = 1s; 2/ .b, v = x' =-8 )2/.2sin( t cm/s
a = - 2x = - 16 2 )2/.2cos( t (cm/s2).
c, v=-4 ; a=8 3.2
; V av < 0 nn chuyn ng chm dn.
V d 3:Cho cc phng trnh dao ng iu ha nh sau:
a. x 5cos 4 t6
(cm). b. x 5cos 2 t4
(cm)
c. x 5cos t (cm). d. x 10sin 5 t3
(cm)
Xc nh A, , , f, T ca cc dao ng iu ha ?Hng dn:
a. x 5cos 4 t6
(cm)
- in : A = 5 (cm). - Tn s gc: = 4 (rad/s).
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- Pha ban u: rad6
.- Tn s: 42 f f 2 Hz
2 2
.- Chu k:
1 1T 0,5 s
f 2 .
b. x 5cos 2 t4
(cm)
V bin A > 0 nn phng trnh dao ng iu ha c vit li:
5x 5cos 2 t 5cos 2 t
4 4
(cm)
- in :A = 5 (cm). - Tn s gc: = 2 (rad/s).
- Pha ban u: 5
rad4
.- Tn s:
2f 1 Hz
2 2
.
- Chu k: 1 1
T 1 sf 1
.
c. x 5cos t 5cos t (cm)- in : A = 5 (cm). - Tn s gc: = (rad/s).
- Pha ban u: rad .- Tn s: f 0,5 Hz2 2
.
- Chu k: 1 1
T 2 sf 0,5
.
d. x 10sin 5 t3
(cm)
- in : A = 10 (cm). - Tn s gc: = 5 (rad/s).
- Pha ban u: rad3
. - Tn s:
5f 2,5 Hz
2 2
. - Chu k:
1 1T 0,4 s
f 2,5 .
V d 4: Mt vt chuyn ng trn u vi tc gc l rad/s. Hnh chiu ca mt vt trn mtng knh daong iu ha vi tn s gc, chu k v tn s bng bao nhiu?
A. rad/s; 2s; 0,5Hz B. 2 rad/s; 0,5s; 2Hz
C. 2 rad/s; 1s; 1Hz D.2
rad/s; 4s; 0,25Hz
Hng dn:Ta c: 2 1
/ 2 0,5rad s T s f Hz T
V d 5: Mt vt dao ng iu ha c qu o l mt on thng di 12cm. in dao ng ca vtl:
A. 12cm B. -12cm C. 6cm D. -6cm.
Hng dn:Ta c:12
2 62 2
ll A A cm
V d 6:Chn phng n ng.
Phng trnh ca mt vt dao ng iu ha c dng: )()6
cos(6 cmtx
.Dao ng c:
A. Chu k dao ng l 2 s, bin 6 cm v pha ban u .6
B. Chu k dao ng l 2 s, bin 6 cm v pha ban u .6
5
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C. Chu k dao ng l 0,5s, bin - 6 cm v pha ban u6
D. Chu k dao ng l 2 s, bin - 6 cm v pha ban u6
Hng dn : T phng trnh dao ng ta c: 6cos( ) 6cos( )6 6
x t cm t cm
A = 6 cm;5
/ ;6
rad s
.p n: B
V d 7: Mt cht im dao ng iu ha theo phng trnh: 2 2 os 24
x c t cm
. Dao ng iu ha
ng vi mt chuyn ng trn u c c im sau:
A. Tc gc chuyn ng trn l 2 /rad s .Ta gc ban u ca vtchuyn ng trn l4
C. ng knh qu o trn l 4 2cm D. A, , C u ng
Hng dn: 2 2 os 2 2 2 ; 2 / ;4 4x c t cm A cm rad s
Tc gc chuyn ng trn l 2 /rad s A ngng knh qu o l 2 2 4 2d R A cm cm C ng
Ta gc ban u ca vt chuyn ng trn l4
B ng
V d 8: Cho cc chuyn ng c m t bi cc phng trnh sau:a) 5. ( . ) 1x cos t (cm)
b) 22.sin (2. . )
6
x t
(cm)
c) 3.sin(4. . ) 3. (4. . )x t cos t (cm)Chng minh rng nhng chuyn ng trn u l nhng dao ng iu ho dng sin. Xc nh bin , tn s,
pha ban u, v v tr cn bng ca cc dao ng .Hng dn:
a) 5. ( . ) 1x cos t (cm) 1 5. ( . ) 5.sin( . )2
x cos t t
. (cm)
t x-1 = X. ta c: 5.sin( . )2
X t
(cm) l mt dao ng iu ho
Vi 5( ); 0,5( ); ( )
2. 2. 2
A cm f Hz Rad
VTC ca dao ng l : 0 1 0 1( ).X x x cm
b) 22.sin (2. . ) 1 (4. . ) 1 sin(4. . ) 1 sin(4. . )6 3 3 2 6
x t cos t t t
t X = x-1 sin(4. . )6
X t
l mt dao ng iu ho.
Vi4.
1( ); 2( ); ( )2. 2. 6
A cm f s Rad
c) 3.sin(4. . ) 3. (4. . ) 3.2sin(4. ). ( ) 3. 2.sin(4. . )( )4 4 4
x t cos t t cos x t cm
l mt dao ng iu ho. Vi 4.3. 2( ); 2( ); ( )2. 4
A cm f s Rad
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C.BI TP TLUYN DNG 1:
Cu 1 :i vi dao ng tun hon, khong thi gian ngn nht sau trng thi dao ng lp li nh c gi l :
A. Tn sdao ng. B. Chu k dao ng. C. Pha ban u. D. Tn sgc.
Cu 2:Trong phng trnh dao ng iu ha x = Acos( t ), i lng ( t ) gi l
A. in ca dao ng. B. Tn sgc ca dao ng.
C. Pha ca dao ng. D. Chu k ca dao ng
Cu 3 :i vi mt dao ng iu ho th nhn nh no sau y l sai ?
A.Li bng 0 khi vn tc bng 0. B.Vn tc bng 0 khi lc hi phc ln nht.
C.Vn tc bng 0 khi thnng cc i. D.Li bng 0 khi gia tc bng 0
Cu 4 :Gia tc ca cht im dao ng iu ha bng khng khi vt c
A. Li cc i. B. Vn tc cc tiu C. Li bng khng . D.Vn tc bng khng.
Cu 5 :Vt dao ng iu ha theo phng trnh x = - 5cos( t +6 ). Pha ban u ca dao ng .
A. = /6. B. = - /6. C. = -5 /6 D. = 5 /6
Cu 6 :Phng trnh dao ng ca mt vt dao ng iu ha c dng x = 6cos(10 t + ). Cc n vsdng lcentimet v giy. Ly dca vt khi pha dao ng bng300l
A.3
3 B. 23 . C. 3 3 . D. 32
Cu 7:Mt vt dao ng iu ho c pt l: x = Acos t .
Gc thi gian t = 0 c chn lc vt vtr no di y.
A.Vt qua VTCB theo chiu dng quo B.Vt qua VTC ngc chiu dng quo
C.Khi vt vtr bin dng D. Khi vt vtr bin m
Cu 8 :Trong dao ng iu ha x = Acos( t ), gia tc bin i iu ha theo phng trnh
A. a = Acos( t ). B. a = A 2 cos( t ).
C. a = - A 2 cos( t ). D. a = - Acos( t ).
Cu 9 :Trong dao ng iu ha, li , vn tc v gia tc l ba i lng bin i iu ha theo thi gian v c :
A. Cng bin . B. Cng pha. C. Cng tn sgc. D. Cng pha ban u.
Cu 10 :Cho dao ng iu ha c phng trnh: x = Acos( t + ) trong A, ,l cc hng s. Chn cung trong cc cu sau:
A. i lng gi l pha dao ng.
B. A khng phthuc vo v , n chphthuc vo tc dng ngoi lc kch thch ban u ln hdao ng.
C. i lng gi l tn sdao ng, khng phthuc vo cc c im ca hdao ng.
D. Chu k dao ng c tnh bi T = 2.
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Cu 11:Mt cht im thc hin dao ng iu ha vi chu k T = 3,14(s) v bin A = 0,1(m). Khi cht imqua vtr cn bng th vn tc ca cht im l
A. 0,1m/s. B. 0,3m/s. C. 0,2m/s. D. 0,05m/s.
Cu 12:Mt vt dao ng iu ha vi chu k 1,5/ 2 (s). Khong cch gia 2 vtr bin l 12 2 cm.Vn tc cci c gi tr
A. vmax= 18cm/s B. vmax= 12cm/s C. vmax= 15cm/s D. vmax= 4cm/s.
Cu 13:Vt dao ng iu ha vi bin 6 cm, khi i qua vtr cn bng vt c vn tc 24 cm/s. Vt dao ngvi tn s:
A. 0,5Hz B. 2Hz C. 4Hz D. 6Hz.
Cu 14. Mt cht im dao ng iu ha theo phng trnh: 2 2 os 24
x c t cm
. Dao ng iu ha
ng vi mt chuyn ng trn u c c im no sau y l sai:
A. Tc gc chuyn ng trn l 2 /rad s
. Ta gc ban u ca vt chuyn ng trn l 4
C. ng knh qu o trn l 4 2cm D. Chu k ca chuyn ng trn u l 2s
Cu 15:Mt vt dao ng iu ha theo phng trnh x = 6cos(4 t)cm, tn sdao ng ca vt l
A. 6Hz. B. 4Hz. C. 2Hz. D. 0,5Hz
Cu 16:Mt vt dao ng iu ha theo phng trnh : x = 3cos(33
t ) cm. tnh chu k dao ng v li ca
vt lc t = 0.
A.T = 1s; x = 2 3 cm. B.T = 6s; x = 1,5cm C.T = 1s; x = 1,5 3 cm D.T = 6s; x = 2 3 cm
Cu 17:Mt vt dao ng iu ha c phng trnhdao ng l x = 5cos( 2 t +3
), ( x tnh bng cm, t tnh bng
s; ly 2 10, = 3,14). Gia tc ca vt khi c ly x = 3 cm l
A. -12(m/s2). B. -120(m/s2). C. -1,20(m/s2). D. -60(m/s2).
Cu 18:Mt vt dao ng iu ha vi bin A khng i. Khi chu k dao ng l T th gi trcc i ca vn tc
l v0. Nu chu k dao ng gim i 2 ln, th vn tc ca vt c gi trcc i (v0)
A. '0v = v0 2 B.'
0v = 2v0 C.'
0v = v0/ 2 D.'
0v = v0/2
Cu 19:Mt vt dao ng iu ha theo phng trnh x = 6cos(4 t)cm, vn tc ca vt ti thi im t = 7,5s l:
A. 0cm/s. B. 5,4cm/s. C. - 75,4cm/s. D. 6cm.
Cu 20:Mt vt dao ng iu ha gia 2 vtr bin cch nhau 6 3 cm. trong khong thi gian 6s, vt thc hinc 8 dao ng ton phn. Tnh vn tc trung bnh ca vt trong mt chu k( vTB)
A. vTB= 16 3 cm/s B. vTB= 18 3 cm/s C. vTB= 15 3 cm/s D. vTB= 12 3 cm/s
Cu 21: Phng trnh dao ng iu ha l: 2cos 56
x t
cm. in , pha ban u, pha dao ng thi
im t ln lt l:
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A. 2cm ;6
rad; 5
6t
B. 2cm ;6
rad; 5
6t
C. 2cm ; 56
t
;6
rad D. 2cm ; 5
6t
;6
rad.
Cu 22: Mt vt dao ng iu ha theo phng trnh 5cosx t cm . Tc ca vt c gi tr cc i l:
A. -5 cm/s B. 5 cm/s C. 5 cm/s D.5
cm/s
Cu 23: Trong dao ng iu ha:A. Gia tc bin i iu ha cng pha vi li . . Gia tc bin i iu ha ngc pha vi li .
C. Gia tc bin i iu ha sm pha2
vi li . D. Gia tc bin i iu ha tr pha
2
vi li .
Cu 24:Mt vt dao ng iu ha theo phng trnh 6cos4x t cm . Tn s dao ng ca vt l:A. 6Hz B. 4Hz C. 2Hz D. 0,5Hz
Cu 25: Mt cht im dao ng iu ha theo quy lut: 2 2 os 24
x c t cm
. Dao ng iu ha ng
vi mt chuyn ng trn u c c im sau:
A. n knh qu o l 4cm . Chu k quay ca chuyn ng trn l 2 2s C. Tc di ca chuyn ng trn u l 4 2 /cm s D. A, , C u sai
Cu 26: Mt cht im dao ng iu ho theo phng trnh: cmtx )3
2cos(4
, bin dao ng ca cht
im l:
A. 4m B. 4cm C.3
2m D.
3
2cm
Cu 27: Cho phng trnh dao ng : 6 os 4 t + ( )6
x c cm
. Pha ca dao ng v li ti thi im1
4t s l:
A. 7 ;3 36rad cm B. 7 ; 3 36
rad cm C. 7 ; 36 rad cm D. 7 ;36 rad cm Cu 28. Mt vt dao ng iu ha tn s 2Hz, bin A = 5 cm. Ly = 10. Khi vn tc ca vt c ln l16 cm/s th gia tc ca vt c ln l:
A.6,4m/s B. 4,8m/s C. 2,4 m/s D. 1,6m/s
Cu 29.Mt vt dao ng iu ho vi vn tc cc i bng 1,64m/s v gia tc cc i bng 65,6m/ . Qu ochuyn ng ca vt l mt on thng c di bng:
A. 8,2 cm B. 4,1 cm C. 5,4 cm D. 10,8 cm
Cu 30. Mt cht im dao ng iu ha theo phng t rnh: 2 2 os 24
x c t cm
. Dao ng iu ha
ng vi mt chuyn ng trn u c c im no sau y l sai:
A. Tc gc chuyn ng trn l 2 /rad s . Ta gc ban u ca vt chuyn ng trn l4
C. ng knh qu o trn l 4 2cm D. Chu k ca chuyn ng trn u l 2sp n& Hng dn chi tit:
Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8 Cu 9 Cu10
B C A C C C C C C B
Cu 11 Cu 12 Cu 13 Cu 14 Cu 15 Cu 16 Cu 17 Cu 18 Cu 19 Cu 20
C D B D C B C A A ACu 21 Cu 22 Cu 23 Cu 24 Cu 25 Cu 26 Cu 27 Cu 28 Cu 29 Cu 30
B B B C D B B B A D
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Cu 1:B
Cu 2:C
Cu 3:A
Cu 4:C
Cu 5: Ta c:5
5cos 5cos 5cos6 6 6
x t t t
Pha ban u: =5
6
(rad)
Cu 6:x=6cos(10t+); vi pha dao ng (10t+)= -300x= 3 3 cm
Cu11:T=2
= 2 (rad/s )vmax= A= 0,2m/s
Cu12:T=2
=
2 2.
1,5
; A=
12 2
2; )vmax= A= 4 cm/s
Cu13: vmax= A= 2fA f= 2Hz
Cu 14.Chon D HD: 2 2 os 2 2 2 ; 2 / ;4 4
x c t cm A cm rad s
Tc gc chuyn ng trn l 2 /rad s A ngng knh qu o l 2 2 4 2d R A cm cm C ng
Ta gc ban u ca vt chuyn ng trn l4
B ng. Vy D sai
Cu15:2
f
= 2Hz
Cu16: 3cos( )3 3
x t
T=2
= 6s ; t=0 th x= 1,5cm
Cu17: 5cos(2 )3
x t
vi a= - 2 .x= -1,2m/s2
Cu18:T=2
; vmax= A v
12
2
TT
'
0 1
0 2
v T
v T = 2
Cu19:x=6cos(4 t) v= -24 sin 4 .7,5= 0
Cu20:A=6 3
2= 3 3 ; T =
t
N= 3/4s vtb=
4A
T= 16 3 cm/s
Cu 21: Chn
Cu 22: Chn .HD: Ta c: max 5 /v A cm s Cu 23: Chn .HD: Ta c: os t+x Ac 2 2os t+ os t+a A c A c
Cu 24:Chn C.HD: Ta c: 6cos 4 4 / 22
x t cm rad s f Hz
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Cu 25: Chn D.HD: 2 2 os 2 2 2 ; 2 /4
x c t cm A cm rad s
n knh qu o l 2 2R A cm A sai
Chu k quay ca chuyn ng trn l2 2
22
T s
B sai
Tc di ca chuyn ng trn u l 2.2 2 4 /v R cm s C sai
Cu 26: Chn .HD:Ta c: 24cos( ) 43
x t cm A cm
Cu 27: Chn .HD: Ta c: 6 os 4 t + ( )6
x c cm
Pha ca dao ng ti thi im1
4t s l:
1 74 . +
4 6 6 6
Li ti thi im1
4t s l:
76 os 3 3( )
6x c cm
Cu 28. .Chn .HD:2 4 2 2 2
. .a A v . Th s:
2 4 2 2 2 2 4
2
(4 ) .(5.10 ) (4 ) .(0,16 ) 25600.25.10 160.0, 256 64 40, 96
4,8 /
a
a m s
Cu 29: Chn A.HD : radva
40
max
max , A= cmv 1,4max
cmL 2,8
Cu 30. HD: 2 2 os 2 2 2 ; 2 / ;4 4
x c t cm A cm rad s
Tc gc chuyn ng trn l
2 /rad s A ng
ng knh qu o l 2 2 4 2d R A cm cm C ng
Ta gc ban u ca vt chuyn ng trn l4
B ng.Chon D
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Dng 2: Tnh vn tc, gia tc ca vt dao ng iu ha.
A. Kin thc cn bn:
Vn tc vt ti thi im 0t . 0Asinv t ; Gia tc vt ti thi im 0t . 2
0A osa c t
Vn tc vt ti thi im x.T2
2 2 2 2
2A A
vx v x
; Gia tc vt ti thi im x: 2a x
B. Cc v d:
V d 1: Mt vt dao ng iu ha vi phng trnh: 8 os 103
x c t cm
. Tm li , vn tc v gia tc
ca vt ti thi im 0,2 ?t s
Hng dn chi tit:
T 228 os 10 ' 80 sin 10 10
3 3x c t cm v x t v a x x
Ti 0,1 ?t s 8 os 10 0,1 4 80 os 10 0,1 40 3 /3 3
x c cm v c cm s
2 2 2 210 10 . 4 400 / .a x cm s
V d 2:Mt cht im dao ng iu ha theo phng trnh 6cos 4x t cm, gia tc ca vt ti thi im5t s l:
A. 0cm/s2 B. 947,5cm/s2 C. -947,5cm/s2 D. 94,75cm/s2
Hng dn :Ta c:
2 26cos 4 . 96 os4 t cm/sx t a c
2 2 2 210 96 cos 4 .10 96 / 947,5 /t s x cm s cm s
V d3:Xc nh vn tc v gia tc ca vt dao ng iu ho. Chn cu tr li ng
Mt cht im dao ng iu ha vi phng trnh: )(2
20cos6 cmtx
. thi im st15
vt c:
A.Vn tc scm /360 , gia tc 2/12 sm v ang chuyn ng theo chiu dng qu o.
B. Vn tc scm /360 , gia tc2
/12 sm v ang chuyn ng theo chiu m qu o.C. Vn tc scm /60 , gia tc 2/312 sm v ang chuyn ng theo chiu dng qu o.
D. Vn tc scm /60 , gia tc 2/312 sm v ang chuyn ng theo chiu m qu o.
Hng dn :iu thc vn tc: )/(2
20sin120' scmtxv
Khi st15
: )/(60
6
5sin120
215.20sin120 scmv
0v chuyn ng theo chiu m qu
o
iu thcgia tc: )/(2
20cos2400' 2scmtva
)/
220cos24 2smt
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Khi st15
: 2/312
6
5cos24
215.20cos24 sma
. p n: D
V d4:Cng thc c lp vi thi gianChn cu tr li ng.
Mt vt dao ng iu ha theo phng trnh: )(2
cos4 cmtx
.Vn tc ca vt khi n qua li
cmx 2 l:A. scm /32 B. scm /32
C. C A, u ng D. Mt kt qu khc
Hng dn :Cng thc c lp vi thi gian:2
222
vxA
Vn tc ca vt l: scmxAv /3222 .p n: C
V d5: Mt vt dao ng iu ha phi mt 0,025s i tim c vn tc bng khng ti im tip theo cngc vn tc bng khng, hai im y cch nhau 10cm. Chon p n ng
A.chu k dao ng l 0,025s B.tn sdao ng l 10Hz
C.bin dao ng l 10cm D.vn tc cc i ca vt l 2 /cm s
Hng dn:
ax
2.0,025 0,05( )0,02522
. . 2 /105 0,05
22
m
TT s
v A A m sl TA cm m
A
p n D
V d6: Mt vt dao ng iu ha, thi im t1vt c li x1= 1cm, v c vn tc v1= 20cm/s. n thi im
t2vt c li x2= 2cm v c vn tc v2= 10cm/s. Hy xc nh bin , chu k, tn s, vn tc cc i ca vt?
Hng dn:
Ti thi im t ta c : os( )x Ac t v ' sin ( t+ )v x A ; Suy ra:2
2 2
2
vA x
- Khi t = t1th:2
2 2 11 2
vA x
(1); - Khi t = t2th :
22 2 2
2 2
vA x
(2)
- T (1) v (2)2 2
2 21 21 22 2
v vx x
2 22 2 1
2 2
1 2
100 10( / )v v
Rad sx x
Chu k: T =2
0,628
(s); Tn s: 1,59
2f
Hz; in :
2
201 5
10A
(cm)
Vn tc cc i: Vmax= 10 5A (cm/s)
V d 7:Mt cht im dao ng iu ha theo phng trnh x 6cos 4 t6
, trong x tnh bng cm, t
tnh bng s. Xc nh li , vn tc v gia tc ca cht im khi t 0,25 s .
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Hng dn:Khi t = 0,25 s th:
- Li ca cht im:
3
x 6cos 4 .0,25 6cos 6cos 6. 3 3 cm6 6 6 2
- Vn tc ca cht im:
v x ' Asin t 24 sin 24 .sin 12 37,686 6
(cm/s).
- Gia tc ca cht im:
2 2 2 23
a v' Acos t 16 .6cos 96 . 48 3 820,56 2
(cm/s2).
Hoc: 2 2a x 16 . 3 3 820,5 (cm/s2).
V d 8:Mt vt nh c khi lng 100 g dao ng iu ha trn qu o thng di 20 cm, vi tn s gc 6 rad/s.Tnh tc cc i v gia tc cc i ca vt.
Hng dn:
- in dao ng ca vt: 20
A 10 cm2 2
- Tc cc i ca vt: maxv A 6.10 60 cm/s 0,6 m/s - Gia tc cc i ca vt: 2 2 2 2maxa A 6 .10 360 cm/s 3,6 m/s
V d 9:Mt vt dao ng iu ha trn qu o di 40 cm. Khi vt v tr c li 10 cm vt c vn tc 20 3 cm/s. Tnh tc cc i v gia tc cc i ca vt.
Hng dn :-in dao ng ca vt:
40
A 20 cm2 2 Tm = ?
T h thc c lp vi thi gian: 2
2 2
2 2 2 2 2
v v 20 3x A 2 rad/s
A x 20 10
- Tc cc i ca vt: maxv A 2 .20 40 cm/s - Gia tc cc i ca vt: 2 2 2 2maxa A 4 .20 80 cm/s
V d 10:Mt cht im dao ng iu ha vi chu k 0,314 s v bin 8 cm. Tnh vn tc ca cht im khi nqua v tr cn bng v khi n qua v tr c li 4 cm.
Hng dn :
- Tm = ? 2 2
20 rad/sT 0,314
- Khi vt qua v tr cn bng th vn tcca vt t gi tr cc i:
maxv A 20.8 160 cm/s - Khi vt qua v tr c li x = 4 cm th:
2
2 2 2 2 2 2
2
vx A v A x 20. 8 4 139 cm/s
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C. BI TP TLUYN DNG 2
Cu 1: Trong dao ng iu ha :A. Vn tc bin i iu ha cng pha vi li . . Vn tc bin i iu ha ngc pha vi li
C. Vn tc bin i iu ha sm pha2
vi li . D. Vn tc bin i iu ha tr pha
2
vi li
Cu 2: Mt vt dao ng iu ha vi tn s f th vn tc cc i c gi tr l1
v . Nu chu k dao ng ca vt tng
2 ln th vn tc cc i c gi tr2
v . Mnh no sau y ng:
A. 1 22v v B. 1 22v v C. 2 12v v D. 2 12v v Cu 3: Mt vt nh chuyn ng trn u theo mt qu o tm O, bn knh R. Trong 12s vt quay c 18 vng.
Gi P l hnh chiu vung gc ca vt trn trc tung. it bn knh qu o trn l 3 2cm ; ly 2 10 . S ovn tc cci v gia tc cc i chuyn ng ca P l:
A. 29 2 / ;270 2 /cm s cm s B. 28 2 / ;240 2 /cm s cm s C. 29 2 / ;270 /cm s cm s D. 28 2 / ;240 /cm s cm s
Cu 4: Mt vt dao ng iu ha vi tn s 1Hz. Lc 0t , vt qua v tr M m 3 2Mx cm vi vn tc
6 2 /cm s . in ca dao ng l:A. 6cm B. 8cm C. 4 2cm D. 6 2cm
Cu 5: Trong dao ng iu ha, ln cc i ca vn tc l:
A. maxv A B.2
maxv A C. maxv A D.2
maxv A Cu 6: Cho mt vt dao ng iu ho vi phng trnh x = 10cos(10t) cm. Vn tc ca vt c ln 50cm/s lnth2012 ti thi im
A.60
6209s B.
12
1207s C.
12
1205s D
60
6031s
Cu 7: Mt vt nhdao ng iu ha vi chu kT=1s. Ti thi im t1no , li ca vt l -2cm. Ti thi imt2 = t1+0.25s,vn tc ca vt c gi tr:A: 4cm/s B:-2m/s C:2cm/s D:- 4m/sCu 8: Mt vt dao ng iu ha phi mt 0,025s i tim c vn tc bng khng ti im tip theo cng cvn tc bng khng, hai im y cch nhau 10cm. Chon p n ng
A.chu k dao ng l 0,025s B.tn sdao ng l 10Hz
C.bin dao ng l 10cm D.vn tc cc i ca vt l 2 /cm s
Cu 9: Mt vt dao ng iu ho theo phng trnh : x = 10 cos (3
4
t ) cm. Gia tc cc i vt l
A. 10cm/s
2
B. 16m/s
2
C. 160 cm/s
2
D. 100cm/s
2
Cu 10.Mt vtdao ngiu ha cphng trnh : x 2cos(2t /6) (cm, s) Li v vn tc cavtlc t
0,25s l :
A. 1cm ; 2 3 .(cm/s). . 1,5cm ; 3 (cm/s). C. 0,5cm ; 3 cm/s. D. 1cm ; cm/s.
Cu 11.Mt vtdao ngiu ha cphng trnh : x 5cos(20t /2) (cm, s). Vn tc cc i v gia tc cc
i cavtl :A. 10m/s ; 200m/s2. B. 10m/s ; 2m/s2. C. 100m/s ; 200m/s2. D. 1m/s ; 20m/s2.
Cu 12.Vtdao ngiu ha theo phng trnh : x 10cos(4t +8
)cm. it li cavtti thi im t l
4cm. Li cavtti thi im sau 0,25s l :A. -4cm B. 10cm C. 4 2cm D. 6 2cm
Cu 13:Mt vt dao ng iu ha theo phng trnh: 3cos(2 )3
x t
, trong x tnh bng cm, t tnh bng
giy. Gc thi gian c chn lc vt c trng thi chuyn ng nh thno?
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A. i qua Vtr c li x = - 1,5 cm v ang chuyn ng theo chiu dng trc OxB. i qua vtr c li x = 1,5 cm v ang chuyn ng theo chiu m ca trc OxC. i qua vtr c li x = 1,5 cm v ang chuyn ng theo chiu dng trc OxD. i qua vtr c li x = - 1,5cm v ang chuyn ng theo chiu m trc Ox
Cu 14. Mt vt dao ng iu ha vi chu k T = 1s. thi im pha dao ng l rad4
3vn tc ca vt c gi
tr l v = - 4 2 cm/s. Ly 2= 10. Gia tc ca vt thi im cho nhn gi tr no?
A.0,8 2 m/s2 B. -0,8 2 m/s2 C.0,4 2 m/s2 D.-0,4 2 m/s2Cu 15. Mt con lc l xo gm l xo c cng 20 N/m v vin bi c khi lng 0,2 kg dao ng iu ha. Ti
thi im t, vn tc v gia tc ca vin bi ln lt l 20 cm/s v 22 3 /m s . in dao ng ca vinbi l
A. 4 cm. B. 16 cm. C. 10 3cm D. 4 3cm
p n & Hng dn chi tit:
Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8 Cu 9 Cu 10
A A A A A B A D B A
Cu 11 Cu 12 Cu 13 Cu 14 Cu 15 Cu 16 Cu 17 Cu 18 Cu 19 Cu 20
D A C A A
Cu 1: HD: Ta c: os t+x Ac
sin t+ sin t+ os t+ os t+2 2
v A A A c A c
Cu2: HD: 1 1Tf ; 1 1 1
2v A AT ; 2 12T T ; 12 2
2 1
2 22 2
vv A A AT T 1 22v v
Cu 3: HD: Chu k: 2 2 2 .3
3 /3 2
T s rad sT
Vn tc cc i: ax2
. 3 2.3 9 2 /mv A A cm sT
Gia tc cc i: 2 2 2ax 9 .3 2 270 2 /ma A cm s
Cu 4: HD:Ta c: 2 2 /f rad s ;2 2
2 2
2 2
2.3618 36 6
4
MM
vA x A cm
Cu 5: HD:Ch n A. maxv A
Cu 6: Gii: chu k : T =
10
2= 0,2 (s);x = 10cos(10t) cm => v = x = - 100sin10t (cm/s)
v = 50cm/s => sin10t = 0,5 => x = 5 3 cm
Trong mt chu k c 4 ln vt c ln vn tc bng 50cm/s
Khi t = 0 vt bin dng. Nn ln th2012 vt c ln vn tc bng 50cm/s khi x = 5 3 cm
v ang chuyn ng theo chiu dng vbin dng vo thi im:
t = (2012:4) TT/12 = (503 -12
1) T =
12
6035.0,2 =
12
1207(s). Chn B
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Cu 7: Gii: Gisphng trnh dao ng ca vt c dng x = AcosT
2t (cm)
x1= AcosT
2t1 (cm)
x2= AcosT
2t2= Acos
T
2(t1+
4
T) = Acos(
T
2t1 +
2
) (cm) = - Asin
T
2t1
v2= x2= -T
2Asin(
T
2t1 +
2
) = -
T
2Acos
T
2t1= 4(cm/s).Chn A
Cu 8: Ch n D.Gii:ax
2.0,025 0,05( )0,02522
. . 2 /105 0,05
22
m
TT s
v A A m sl TA cm m
A
Cu 9: Ch n . Gia tc cc i: 2 2 2max (4 ) . 160.10 16 /a A A m s
Cu 10.HD : T phng trnh x 2cos(2t /6) (cm, s) v 4sin(2t /6) cm/s.
Thay t 0,25s vo phng trnh x v v, ta c : x 1cm, v 2 3 (cm/s) Chn : A.
Cu 11.HD :p dng : maxv A v maxa 2A Chn : D
Cu 12.HD : Ti thi im t : 4 10cos(4t +/8)cm. t : (4t +/8) 4 10cos
Ti thi im t +0,25: x 10cos[4(t +0,25) +/8] 10cos(4t +/8 +) 10cos(4t +/8) 4cm.
Vy : x 4cm
Cu 13.HD:0
'
0
3cos 2 .0 1,53
6 sin 2 .0 3 3 / 03
x cm
v x cm s
p n C
Cu 14. Ch n A.3
sin( ) 4 2 2 sin( ) 44
v A t A A cm
2 2 2 2 23 2s( ) (2 ) 4 s( ) (2 ) 4( ) 8 2 /4 2
a Aco t a co cm s
Cu 15: p n A.Ta c:
- Tn s gc: 10 /k
rad sm
-Li ti thi im t: 2 2 2 3a
a x x cm
- in dao ng: 22 2
2
2
202 3 4
10
vA x cm
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Dng 3: L in hx, v va ca vt dao ng iu ha.
A. Kin thc cn bn:
a) T cc phng trnh ca vn tc v li ta c2 2x A cos t x v
1, 1A Av A sin t
(1) c gi l h thc lin h ca x, A, v v khng ph thuc vo thi gian t.
+2
2
2
v1 A x
+2
2
2
v1 x A
+ 2 21 v A x nu v l tc th 2 2v A x
+2 2
v1
A x
+ Vi hai thi im t1, t2vt c cc cp gi tr x1, v1v x2, v2th ta c h thc sau:2 2
1 2
2 22 2 2 2 2 2 2 21 21 1 2 2 1 2 1 2
2 2 22 2
1 2
2 2
1 2
v v
x xx v x v x x v v
A A A A A A x xT 2
v v
b) T cc phng trnh ca vn tc v gia tc ta c2 2 2 2
2 4 2 2 22
v A sin t a v a v1 1, 2
A A A Aa A cos t
(2) c gi l h thc lin h ca a, A, v v khng ph thuc vo thi gian t.
c) Lin h22 2
2, v: vA x v A x
; Lin h
2 22
4 2, a: a vA v v A
.
d) Lc ko v( hay lc hi phc): axKV KVmF kx F KA .
e) ax ax axax3 3
0 ; ; ;2 2 2 22 2
m m mm
v v vA A AKhi x v A Khi x v Khi x v Khi x v
Ch :
+ T (1) ta thy th ca (v, x) l ng elip.+ T (2) ta thy th ca (a, v) l ng elip.+ T a = 2x ta thy th ca (a, x) l on thng.
B. Cc v d:
V d1:Mt vt dao ng iu ha: khi vt c li 1 3x cm . Th vn tc l 1 4 /v cm s , khi vt c li
2 4x cm th vn tc l 2 3 /v cm s . Tm tn sgc v bin ca vt?
Hng dn :
T
222 2 2 21
1 2 2
2 22 2 2 2 2
2 2 2
43
/
534
vA x A
rad s
A cmvA x A
V d 2:Mt vt dao ng iu ha vi phng trnh x = 5cos(t + /3) cm. Ly = 10.
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a) Khi vt qua v tr cn bng c tc 10(cm/s). Vit biu thc vn tc, gia tc ca vt.b) Tnh tc ca vt khi vt c li 3 (cm).
c) Khi vt cch v tr cn bng mt on 5 2
2cm th vt c tc l bao nhiu ?
Hng dn :a) Khi vt qua v tr cn bng th tc ca vt t cc i nn
2 maxmax
v 10v A 10 rad / s
A 5
Khi 2 2 2
v x ' 10sin t3
x 5cos 2 t cm3
a A 4 .5cos 2 t cm / s3
b) Khi x = 3 cm, p dng h thc lin h ta c2 2
2 2 2 2x v 1 v 2 A x 2 5 3 8 cm / sA A
c) Khi vt cch v tr cn bng mt on 5 2
2
cm , tc l
2
25 2 5 2x cm v 2 5 5 2 cm / s2 2
V d3:Mt vt dao ng iu ha c 2max max
v 16 cm / s ;a 6, 4 m / s
a) Tnh chu k, tn s dao ng ca vt.b) Tnh di qu o chuyn ng ca vt.
c) Tnh tc ca vt khi vt qua cc li A A 3
x ; x2 2
.
Hng dn :
a) Ta c max max2 2
max max
v 16 cm / s a 640 404 rad / s
a 6, 4 m / s 640 cm / s v 16
.
T ta c chu k v tn s dao ng l
2T 0, 5 s
f 2 Hz2
b) Bin dao ng A tha mn maxv 16
A 4 cm
4
di qu o chuyn ng l 2A = 8 (cm).
c) p dng cng thc tnh tc ca vt ta c:2
2 2 2
22 2 2
A A 4 A 3khi x v A x 4 A 8 3 cm / s
2 4 2
A 3 3A 4 Akhi x v A x 4 A 8 cm / s
2 4 2
V d 4:Mt vt dao ng iu ha theo phng ngang vi bin 2 cm v chu k l 0,2 s. Tnh ln gia tc
ca vt khi n c vn tc 10 10 cm/s. Ly 2 10 .Hng dn :
Ta c: 2 2 10 rad/sT 0,2
Ta chng minh cng thc:2 2
2
2 4
v aA
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Gi s vt dao ng iu ha theo phng trnh x Acos t th:
2 2 2 22 2 2 2
22 2 4 2 2 2 2 2
2
v A sin t (1)v Asin t v A sin t
aa Acos t a A cos t A cos t (2)
Ly (1) cng (2), ta c:
2
2 2 2 2 2 2 2 2 2 2 2 2
2
a
v A a A v a A v 10 100 .2 1000
2 2 2a 10 2000 1000 10 10 10 100 1000 cm/s 10 m/s
V d 5:Mt mt c khi lng m = 100 g dao ng iu ha theo phng trnh x 5cos 2 t6
(cm). Ly
2 10 . Xc nh li , vn tc, gia tc, lc hi phc trong cc trng hp sau:a. thi im t = 5 s.
b. Pha dao ng l 1200.Hng dn :a. thi im t = 5 s.
- Li : 3x 5cos 2 .5 5cos 5. 2,5 3 cm6 6 2
- Vn tc: v 10 sin 2 .5 10 sin 5 cm/s6 6
- Gia tc: 2 2 2a x 4 .2,5 3 100 3 cm/s - Lc hi phc: 2 2hpF m x 0,1.40.2,5 3.10 0,1 3 N
b. Khi pha dao ng 1200. 0120 t2 6
- Li : x 5cos 5sin 2,5 cm2 6 6
- Vn tc: v 2 .5sin 10 cos 5 3 cm/s2 6 6
- Gia tc: 2 2a x 40. 2,5 100 cm/s - Lc hi phc: 2 1 2hpF m x 10 .4.10. 2,5.10 0,1 N
V d 6: Mt cht im dao ng iu ha trn trc Ox. Khi cht im i qua v tr cn bng th tc ca n l 20
cm/s. Khi cht im c tc l 10 cm/s th gia tc ca n c ln l 40 3 cm/s2. in dao ng ca chtim l bao nhiu ?
Hng dn :
- Khi cht im qua VTC th tc ca n t gi tr cc i maxv A 20 cm/s - bi cho: khi v 10 cm/s th 2a 40 3 cm/s
T cng thc:2 2 2 2
2 2 2 2 2 2
max2 4 2 2
v a a aA A v v v
2
2 2
2 2 2 2 2
max
40 3a 3.40
4 rad/sv v 20 10 3.10 m: max
max
v 20
v A A 5 cm4
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C.BI TP TLUYN DNG 3
Cu 1:Vt dao ng iu ho xung quanh v tr cn bng vi chu k T = /15s vi bin A = 5cm. t trc to
Ox nm ngang, gc to O ti v tr cn bng chiu dng sang tri. qua khi lng ca l xo. Vt chuynng theo chiu dng c tc 90cm/s khi i qua v tr c li :
A. x = 3cm B. x= 4cm C.x = -3cm D. x= -2cmCu 2. Mt vt dao ng iu ha vi tn s f = 2Hz. Ly 2= 10. Ti mt thi im t vt c gia tc a = 1,6 m /s2
v vn tc l v = 4 3 cm/s. in dao ng ca vt c gi tr l:
A. 2cm B. 4cm C. 2 3 cm D. 8cm
Cu 3.Mt vt dao ng iu ho, gia tc ca vt ti bin c ln l 8m/ . Khong thi gian vt qua v tr cn
bng 5 ln lin tip l 1s. Ly = 10. in dao ng bng:A.3,2cm B. 4cm C. 5cm D. 10cm
Cu 4.Mt vt dao ng iu ho theo phng trnh x = 2cos(4t) cm. Li v vn tc ca vt thi im t = 0,25
(s) l
A. x =1 cm; v = 4cm/s. B. x =2 cm; v = 0 cm/s.
C. x = 1 cm; v = 4cm/s. D. x = 2 cm; v = 0 cm/s.
Cu 5.Mt vt dao ng iu ho theo phng trnh x = 4sin(5t /6) cm. Vn tc v gia tc ca vt thi im t
= 0,5 (s) l
A. 2 210 3 cm / s; 50 cm / s B. 2 210 cm / s;50 3 cm / s
C. 2 210 3 cm / s;50 cm / s D. 2 210 cm / s; 50 3 cm / s
Cu 6.Mt vt dao ng iu ha c phng trnh x = 5cos(2t /6) cm. Vn tc ca vt khi c li x = 3 cm l
A. v = 25,12 cm/s. B. v = 25,12 cm/s. C. v = 12,56 cm/s D. v = 12,56 cm/s.
Cu 7.Mt vt dao ng iu ha c phng trnh x = 5cos(2t /6) cm. Ly 2
= 10. Gia tc ca vt khi c li x = 3 cm l
A. a = 12 m/s2 B. a =120 cm/s2 C. a = 1,20 cm/s2 D. a = 12 cm/s2
Cu 8.Mt cht im dao ng iu ha vi phng trnh x = 20cos(2t) cm. Gia tc ca cht im ti li x = 10
cm l
A. a =4 m/s2 B. a = 2 m/s2 C. a = 9,8 m/s2 D. a = 10 m/s2
Cu 9( C 2013):Mt vt nh dao ng iu ha vi bin 5 cm v vn tc c ln cc i l 10 cm/s.Chu k dao ng ca vt nh lA. 4 s. B. 2 s. C. 1 s. D. 3 s.
Cu 10:Mt vt nh chuyn ng trn u theo mt qu o tm O, bn knh R. Trong 12s vt quay c 18 vng.Gi P l hnh chiu vung gc ca vt trn trc tung. it bn knh qu o trn l 3 2cm ; ly 2 10 . Chu k,tn s, s o vn tc cc i v gia tc cc i chuyn ng ca P l:ca chuyn ng ca P l:
A.2
; 1,53
s Hz; 29 2 / ;270 2 /cm s cm s B. 21,5 ;3
s Hz ; 28 2 / ;240 2 /cm s cm s
C. 1,5 ;3s Hz; 29 2 / ;270 /cm s cm s D. 2 ;0,5s Hz; 28 2 / ;240 /cm s cm s
p n & Hng dn chi tit:
Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8 Cu 9 Cu 10
B A C B D B B A C A
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Hng dn chi tit:
Cu 1: sradT
/302
, xvAx 1622
22
4 cm Chn B
Cu 2:Chn A
Cu 3.Gii:5 ln lin tip vt qua VTCB = 1s 2,5T = 1 T = 0,4s = 5 rad/s A = cma 2,32max
Chn CCu 4.Chn B
Cu 5.Chn D
Cu 6.Chn B
Cu 7.Chn B
Cu 8.Chn A
Cu 9.Chn CGii 1: maxmax
2 2 2 .51
10
A Av A T s
T v
. p n C.
Gii 2:vmax= A= maxvA = 2 rad/sT = 2 = 1 s. p n C.
Cu 10:Chn A. HD: Khi vt nh chuyn ng trn u th P dao ng iu ha vi chu k v tn s bng ng
chu k v tn s ca chuyn ng trn:12 2 1
; 1,518 3
T s f Hz T
.
Phng trnh dao ng iu ha: os t+x Ac vi 2 2 2 .3 3 /3 2
T s rad sT
Vn tc cc i: ax2
. 3 2.3 9 2 /mv A A cm sT
Gia tc cc i: 2 2 2
ax 9 .3 2 270 2 /ma A cm s
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Dng 4: Vit phng trnh ca vt dao ng iu ha.
A. Kin thc cn bn:
* Chn hquy chiu : -Trc Ox -Gc ta ti VTCB
-Chiu dng .-Gc thi gian
* Phng trnh dao ng c dng : x Acos(t +) cm
* Phng trnh vn tc : v -Asin(t +) cm/s
* Phng trnh gia tc : a -2Acos(t +) cm/s2
1Tm * cho : T, f, k, m, g, l0
-
2f
2
T
, vi T t
N
, N
Tng sdao ng trong thi gian t
Nu l con lc l xo :
Nm ngang Treo thng ng
k
m, (k : N/m ; m : kg)
0
g
l, khi cho l0
mg
k
2
g
.
cho x, v, a, A : 2 2
v
A x
a
x
maxa
A
maxv
A
2Tm A
* cho : cho x ng vi v A = 2 2vx ( ) .
-Nu v 0 (bung nh) A x
-Nu v vmaxx 0 A maxv
* cho : amax A max
2
a
* cho : chiu di qu o CD A =
CD
2.
* cho : lc FmaxkA. A = maxF
k * cho : lmax v lminca l xo A = max min
l l
2
.
* cho : W hoc dmaxW hoc tmaxW A =2W
k.Vi W WmaxWtmax
21 kA2
.
* cho : lCB,lmax hoc lCB, lmim A =lmaxlCB hoc A =lCBlmin.
3 -Tm (thng ly
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* Nu t 0 : -x x0 , v v0 0
0
x Acos
v A sin
0
0
xcos
A
vsin
A
?
-v v0; a a 0
20
0
a A cos
v A sin
tan 0
0
v
a ?
c bit:+x0 0, v v0(vt qua VTCB)0
0 Acos
v A sin
0
cos 0
vA 0
sin
0
2
vA / /
+x x0, v 0 (vt qua VT bin ) 0x A cos
0 A sin
0xA 0cos
sin 0
o
0;
A /x /
* Nu t t1:1 1
1 1
x Acos( t )
v A sin( t )
?hoc
21 1
1 1
a A cos( t )
v A sin( t )
?
Lu :Vt i theo chiu dng th v >0 sin
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Tm A: ax 5 ; m : 0; 0; 0mv
A cm T t x v
cos 0 os 05cos 2
sin 0 2 2Asin 0
x A t cvay x t cm
v t
V d 2:Mt con lc l xo dao ng vi bin A = 5 cm vi chu k T = 0,5 s. Vit phng trnh dao ng ca conlc trong cc trng hp sau:a. Lc t = 0, vt qua v tr cn bng theo chiu dng.
b. Lc t = 0, vt v tr bin.c. Lc t = 0, vt c li 2,5 cm theo chiu dng.
Hng dn gii:Phng trnh dao ng iu ha ca vt c dng: x Acos t Phng trnh vn tc l: v Asin t
a. Lc t = 0, vt qua v tr cn bngtheo chiu dng.
2 2 4 rad/sT 0,5
Chn t = 0 lc x = 0 v v > 0, khi :0 Acos cos 0
Asin 0 sin 0 2
Vy phng trnh dao ng iu ha ca vt l: x 5cos 4 t2
(cm)
b. Lc t = 0, vt qua v tr c li 5 cm theo chiu dng. Trng hp 1: Vt v tr bin dng.
Chn t = 0 lc x = A v v = 0, khi :5 5cos cos 1
0Asin 0 sin 0
Vy phng trnh dao ng iu ha ca vt l: x 5cos 4 t (cm) Trng hp 2: Vt v tr bin m.
Chn t = 0 lc x = A v v = 0, khi :5 5cos cos 1
Asin 0 sin 0
Vy phng trnh dao ng iu ha ca vt l: x 5cos 4 t (cm)c. Lc t = 0, vt c li 2,5 cm theo chiu dng.
Chn t = 0 lc x = 2,5 cm v v > 0, khi :
12,5 5cos cos
2Asin 0 3
sin 0
Vy phng trnh dao ng iu ha ca vt l: x 5cos 4 t3
(cm)
V d3: Mt cht im dao ng iu ho dc theo trc quanh VTC O vi bin 4 cm, tn s f=2Hz .hylp phng trnh dao ng nu chn mc thi gian t0=0 lca. cht im i quali x0=2 cm theo chiu dng
b. cht im i qua li x0= -2 cm theo chiu m
Hng dn gii:a. t0=0 th30sin.4.4
cos42
0
0
v
x => x=4cos(4 )
3.
t cm
b. . t0=0 th3
.2
0sin.4.4
cos42
0
0
v
x
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V d 4:Con lc l xo gm qu cu c khi lng 300 g, l xo c cng 30 N/m treo vo mt im c nh. Chngc ta v tr cn bng, chiu dng hng xung, gc thi gian l lc vt bt u dao ng. Ko qu cuxung khi v tr cnbng 4 cm ri truyn cho n mt vn tc ban u 40 cm/s hng xung. Vit phng trnh daong ca vt.
Hng dn gii:
Phng trnh dao ng ca vt c dng: x Acos t Phng trnh vn tc ca vt: v Asin t
Ta c: k 30 10 rad/sm 0,3
Tm A = ?
T h thc c lp: 2 2 2
2 2 2 2
2 2 2
v v 40A x A x 4 4 2 cm
10
Chn t = 0 lc x = 4 cm v v = 40 cm/s, khi :
2cos
4 4 2cos 2
440 40 2 sin 2sin
2
Vy phng trnh dao ng ca vt l: x 4 2cos 10t4
(cm)
V d5: Mt cht im d hdc theo trc Ox quanh v tr cn bng O vi srad/10 a. Lp phng trnh dao ng nu chn mc thi gian t0=0 lc cht im i qua li x0=-4 cm theo chiu m vivn tc 40cm/s
b. Tm vn tc cc i ca vt
Hng dn gii: a. t0=0 th
A
A
Av
Ax
4sin
4cos
0sin..1040
cos4
0
0
suy ra 24,4
A
cm
b. vmax= 2.402.4.10. A
V d6: Mt con lc l xo dao ng vi chu k T = 1(s). Lc t = 0, vt qua v tr c li 5. 2x (cm) vi vntc 10. . 2v (cm/s). Vit phng trnh dao ng ca con lc.
Hng dn gii:Phng trnh dao ng c dng : . s( . )x A co t .
Phng trnh vn tc c dng : ' . .sin( . )v x A t .
Vn tc gc :2. 2.
2 ( / )1
Rad sT
.
ADCT :2
2 2
2
vA x
2 22 2
2 2
( 10. . 2)( 5 . 2)
(2. )
vA x
= 10 (cm).
iu kin ban u : t = 0 ;. s
. .sin
x A co
v A
5. 2 . s
10. . 2 .2. .s
A co
A in
tan 1 3.
( )4
rad
. Vy3
10. s(2. . )4
x co t
(cm).
V d 7:Vt dao ng iu ha vi tn s f = 0,5 Hz. Ti t = 0, vt c li x = 4 cm v vn tc v = +12,56 cm/s.Vit phng trnh dao ng ca vt.
Hng dn gii:Phngtrnh dao ng iu ha ca vt c dng: x Acos t Phng trnh vn tc: v Asin t Tm = ?
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Ta c: 2 f 2 .0,5 rad/s Chn t = 0 lc x = 4 cm v v = +12,56 cm/s, khi :
4 Acos Acos 4
Asin 12,56 Asin 4 4
T (1), tasuy ra: 4 4A 4 2 cm
2cos
4 2
=>phng trnh dao ng : x 4 2cos t4
(cm)
V d 8: Mt vt dao ng iu ho dc theo trc Ox. Lc vt qua v tr c li 2x (cm) th c vn tc. 2v (cm/s) v gia tc 22.a (cm/s2). Chn gc to v tr trn. Vit phng trnh dao ng ca vt
di dng hm s cosin.Hng dn gii: Phng trnh c dng : x = A.cos( .t ). Phng trnh vn tc : v = - A. .sin( . )t .
Phng trnh gia tc : a= - A. 2. ( . )cos t .Khi t = 0 ; thay cc gi tr x, v, a vo 3 phng trnh ta c :
2 22 . ; . 2 . .sin ; . 2 .x A cos v A a Acos .Ly a chia cho x ta c : ( / )rad s .
Ly v chia cho a ta c :3.
tan 1 ( )4
rad
(v cos< 0 )
2A cm . Vy :3.
2. s( . )4
x co t
(cm).
V d9: Vt dao ng iu ha vi tc cc i 40 cm/s. Ti v tr c li 0 2 2( )x cm vt c ng nngbng th nng. Nu chn gc thi gian l lc vt qua v tr ny theo chiu dng th phng trnh dao ng ca vtl
Hng dn gii:40
4
4cos 102 10 42 22
AA
x tA
404
4cos 102 10 42 22
AA
x tA
cm
V d 10:Mt con lc l xo dao ng iu ha vi chu k T = 1 s. Lc t = 2,5 s vt qua v tr c li x 5 2 cm v vn tc v 10 2 cm/s. Vit phng trnh dao ng iu ha ca con lc.
Hng dn gii:
Phng trnh dao ng iu ha cdng: x Acos t
Phng trnh vn tc: v Asin t ; Ta c: 2 2 rad/sT
Tm A = ?
2
22
2 2
22
10 2vA x 5 2 50 50 100 A 10 cm
2
Chn t = 2,5 s lc x 5 2 cm v v 10 2 cm/s, khi : 5 2 10cos (1)10 2 20 sin (2)
Ly (2) chia (1), ta c: 2 tan 2 tan 14
Vy phng trnh dao ng iu ha: x 10cos 2 t
4
(cm)
V d 11:Mt vt dao ng iu ha thc hin 10 dao ng trong 5 s, khi vt qua v tr cn bng n c vn tc 20
cm/s. Chn chiu dng l chiu lch ca vt, gc thi gian lc vt qua v tr c li x 2,5 3 cm v angchuyn ng v v tr cn bng. Vit phng trnh dao ng vphng trnh vntcca vt.
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Hng dn gii:
Phng trnh dao ng ca vt c dng: x Acos t Phng trnh vn tc ca vt: v Asin t
Chu k dao ng ca vt: t 5
T 0,5 sn 10
;Tn s gc ca vt: 2 2
4 rad/sT 0,5
Khi vt qua vtr cn bng th vn tc ca vt cci nn: max
max
v 20
v A A 5 cm4
V chiu dng l chiu lch ca vt nn lc t = 0 vt qua v tr x 2,5 3 cm th v < 0.
Khi :3
2,5 3 5cos cos2
6A sin 0sin 0
=>Phng trnh dao ngl: x 5cos 4 t6
(cm)
Phng trnh vntcl: 20 sin(4 )( / )
6
v t cm s
C.BI TP TLUYN DNG 4
Cu 1: Mt vtdao ngiu ha vi bin A = 4cm v T = 2s. Chn gc thi gian l lc vtqua VTCB theochiu dng caqu o. Phng trnhdao ngcavtl :
A. x = 4cos(2t - /2)cm. B. x = 4cos(t - /2)cm.C. x = 4cos(2t -/2)cm. D. x = 4cos(t + /2)cm.
Cu 2: Mt vtdao ngiu ha trn on thng di 4cm vi f = 10Hz. Lc t = 0 vtqua VTC theo chiu m
caqu o. Phng trnh dao ng ca vt l :A. x = 2cos(20t - /2)cm. B. x = 2cos(20t + /2)cm.C. x = 4cos(20t -/2)cm. D. x = 4cos(20t + /2)cm.
Cu 3: Mt l xo u trn c nh, u di treo vtm. Vtdao ngtheo phng thng ng vi tn sgc =10(rad/s). Trong qu trnh dao ng di l xo thay i t 18cm n 22cm. Chn gc ta O ti VTC. Chiudng hng xung, gc thi gian lc l xo c di nh nht. Phng trnh dao ngcavtl :
A. x = 2cos(10t + )cm. B. x = 2cos(0,4t)cm.C. x = 4cos(10t + )cm. D. x = 4cos(10t + )cm.
Cu 4: Mt vtdao ngiu ha vi bin A 4cm v T 2s. Chn gc thi gian l lc vtqua VTCB theo
chiu dng caqu o. Phng trnh dao ngcavtl :
A. x 4cos(2t /2)cm. B. x 4cos(t /2)cm. C. x 4cos(2t /2)cm. D. x 4cos(t /2)cm.Cu 5: Mt vtdao ngiu ha trn on thng di 4cm vi f 10Hz. Lc t 0 vtqua VTC theo chiudng caqu o. Phng trnh dao ng ca vt l :
A. x 2cos(20t /2)cm. B.x 2cos(20t /2)cm. C. x 4cos(20t /2)cm. D. x 4cos(20t /2)cm.
Cu 6: Mt l xo u trn c nh, u di treo vtm. Vtdao ngtheo phng thng ng vi tn sgc10(rad/s). Trong qu trnh dao ng di l xo thay i t 18cm n 22cm. Chn g ta ti VTC. chiudng hng xung, gc thi gian lc l xo c di nh nht. Phng trnh dao ngcavtl :
A. x 2cos(10t )cm. B. x 2cos(0,4t)cm. C. x 4cos(10t )cm. D. x 4cos(10t +)cm.
Cu 7: Mt vt dao ng iu ho trn trc Ox vi tn s f = 4 Hz, bit to ban u ca vt l x = 3 cm v sau
1/24 s th vt li tr v to ban u. Phng trnh dao ng ca vt lA. x = 3 3 cos(8t /6) cm. B. x = 2 3 cos(8t /6) cm.
C. x = 6cos(8t + /6) cm. D. x = 3 2 cos(8t + /3) cm.Cu 8:(H 2013)Mt vt nh dao ng iu ha dc theo trc Ox vi bin 5 cm, chu k 2 s. Ti thi im t =0, vt i qua cn bng O theo chiu dng. Phng trnh dao ng ca vt l
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A. x 5cos( t )2
(cm) B. x 5cos(2 t )
2
(cm) C. x 5cos(2 t )
2
(cm) D. x 5cos( t )
2
Cu 9:(C 2013):Mt vt nhdao ng iu ha dc theo trc Ox (v tr cn bng O) vi bin 4 cm v tn s10 Hz. Ti thi im t = 0, vt c li 4 cm. Phng trnh dao ng ca vt l A. x = 4cos(20t + ) cm. B. x = 4cos20t cm. C. x = 4cos(20t0,5) cm. D. x = 4cos(20t + 0,5) cm.Cu 10:Mt vt dao ng iu ha vi bin A, chu k T. Ti thi im ban u vt c li x = 3cm, chuyn
ng vi vn tc 60 3v cm/s. Sau thi gian mt phn t chu k dao ng vt i qua v tr c li 3 3x cm.Phng trnh dao ng ca vt l
A. 6cos(20 )3
x t cm. B. 6cos(20 )3
x t
cm. C. 6 2 cos(10 )4
x t cm. D. 6 2 cos(10 )x t cm.
Cu 11:Mt vt dao ng iu ho, khong thi gian gia hai ln lin tip vt qua v tr cn bng l 0,5s; qung ng vt i
c trong 2s l 32cm. Ti thi im t=1,5s vt qua li 2 3x cm theo chiu dng. Phng trnh dao ng ca vt l?A:4cos( 2t + /6) cm B:4cos( 2t - 5/6) cm C:4cos( 2t - /6) cm D: 4cos( 2t + 5/6) cm
Cu 12:Mt vt dao ng vi bin 6(cm). Lc t = 0, con lc qua v tr c li x = 3 2 (cm) theo chiu dng
vi gia tc c ln3
2(cm/s2). Phng trnh dao ng ca con lc l:
A. x = 6cos9t(cm) B.t
x 6cos3 4
(cm) C.t
x 6cos3 4
(cm) D. x 6cos 3t3
(cm)
Cu 13:Mt vt dao ng iu ho khi qua v tr cn bng vt c vn tc v = 20 cm/s. Gia tc cc i ca vt lamax= 2m/s
2. Chn t = 0 l lc vt qua v tr cn bng theo chiu m ca trc to . Phng trnh dao ng ca vtl :A. x = 2cos(10t + ) cm. B. x = 2cos(10t + /2) cm. C. x = 2cos(10t/2) cm. D. x = 2cos(10t) cm.Cu 14: Mt vt dao ng iu ho c sau 1/8 s th ng nng li bng th nng. Qung ng vt i c trong0,5s l 16cm. Chn gc thi gian lc vt qua v tr cn bng theo chiu m. Phng trnh dao ng ca vt l:
A. 8 os(2 )2
x c cm
B. 8 os(2 )2
x c cm
C. 4 os(4 )2
x c cm
D. 4 os(4 )2
x c cm
Cu 15:Mt vt dao ng iu ho khi qua v tr cn bng vt c vn tc v = 20 cm/s. Gia tc cc i ca vt lamax= 2m/s
2. Chn t = 0 l lc vt qua v tr cn bng theo chiu m ca trc to . Phng trnh dao ng ca vt
lA. x = 2cos(10t). B. x = 2cos(10t + /2). C. x = 2cos(10t + ). D. x = 2cos(10t/2)Cu 16:Vt dao ng trn qu o di 8 cm, tn s dao ng ca vt l f = 10 Hz. Xc nh phng trnh dao ng ca vt
bit rng ti t = 0 vt i qua v tr x = - 2cm theo chiu m.A:x = 8cos( 20t + 3/4) cm. B:x = 4cos( 20t - 3/4) cm.C:x = 8cos( 10t + 3/4) cm. D: x = 4cos( 20t + 2/3) cm.
Cu 17: Mt vt dao ng iu ho, khong thigian gia hai ln lin tip vt qua v tr cn bng l 0,5s; qung ng vt i
c trong 2s l 32cm. Ti thi im t=1,5s vt qua li 2 3x cm theo chiu dng. Phng trnh dao ng ca vt l?A:4cos( 2t + /6) cm B:4cos( 2t - 5/6) cm C:4cos( 2t - /6) cm D: 4cos( 2t + 5/6) cm
p n & Hng dn chi tit:
Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8 Cu 9 Cu 10
B B A A B A B A B B
Cu 11 Cu 12 Cu 13 Cu 14 Cu 15 Cu 16 Cu 17 Cu 18 Cu 19 Cu 20
D B B D B D D
Cu 1: HD Gii: = 2f = . V A = 4cm loi A v C.
t = 0 : x0= 0, v0> 0 : 0
0 cos
v A sin 0
2sin 0
chn = - /2 Chn: B
Dng My Fx570Esbm: Mode 2, Shift Mode 4(R:radian),
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Nhp:2
4 4cos( )2
, 2 32
4i SHIF x t cmT
Cu 2: HD Gii: = 2f = 20. V A = MN /2 = 2cm loi C v D.
t = 0 : x0= 0, v0< 0 :0
0 cos
v A sin 0
2
sin 0
chn =- /2 Chn: B
Dng My Fx570Esbm: Mode 2, Shift Mode 4(R:radian),
Nhp: 2 4cos( )22
2 32
,2
xi S t cmHIFT
Cu 3: HD Gii: = 10(rad/s) v A = max minl l
2
= 2cm. loi
t = 0 : x0= -2cm, v0= 0 : 2 2cos
0 sin
cos 0
0 ;
chn = x = 2cos(10t + )cm. Chn:A
My Fx570Es : Mode 2, Shift Mode 4(R),Nhp: -2 = 2 3 2 co2
: s( )2SHIFT ketqua x t cm
Cu 4: Gii:2f . v A 4cm loi v D.
t 0 : x00, v0>0 :0
0 cos
v A sin 0
2
sin 0
chn /2 x 4cos(2t /2)cm. Chn : A
Cu 5: Gii:2f . v A MN /2 2cm loi C v D.
t 0 : x00, v0>0 :0
0 cos
v A sin 0
2
sin 0
chn /2 x 2cos(20t /2)cm. Chn :
Cu 6: Gii:10(rad/s) v A max minl l
2
2cm. loi
t
0 : x0
2cm, v0
0 :
2 2cos
0 sin
cos 0
0 ;
chn
x
2cos(10t
)cm.
Chn : ACu 7: Gii :V vng lng gic so snh thi gian cho vi chu k Tsxc nh c v tr ban u ca vt thi im t = 0 v thi im sau 1/24s Ta c: T = 1/f = 1/4s > t = 1/ 24 => vt cha quay ht c mt vngD dng suy ra gc quay = 2 = t = 8/24= /3V cho x = 3cm=> gc quay ban u l = /6
in A = x/ cos= 3/ ( 3 /2) = 2 3 cm=> Chn BCu 8: Gii 1:A= 5cm; =2 /T=2/2 = rad/s.Khi t= 0 vt i qua cn bng O theo chiu dng: x=0 v v>0 => cos = 0 => = -/2 .Chn A.Gii 2:Dng my tnh Fx570ES:Mode 2 ; Shift mode 4:Nhp: -5i = shift 2 3 = kt qu 5 -/2.
Cu 9: Gii 1: Ti thi im t = 0, vt c li x= 4 cm = A , v =0 => =0 . Chn B.Gii 2:= 2f = 20 rad/s; cos=
x
A= 1= 0. p n .
Cu 10: Lu : Nu bi ny da vo p s chn tn s gc ph hp vi= 20rad/s th c th b d d kindo dng my tnh Casio Fx570ES; 570ESPlus Gii nhanh!
(0)
(0)
3
0: 3 3 33 3
a x
t x ivb
. Bm: 3 - 3 3 i = 23 6 6 cos( )3 3
xSHIF tT cm
Chn .
Cu 11: Chn D. Cu 12: Chn B. Cu 13: Chn B.Cu 14: Chn D. Cu 15: Chn B. Cu 16: Chn D.
Cu 17: Chn D.
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Vtr x =2
2A : Wt = W Vtr x =
2
A : W= 3 Wt
Bintri
Binhi
x
T/12
T/4
T/8T/12
T/8
T/4
AOA/2
23A
2A
-A -A/22A 3
2A
T/6T/6
T/12T/12
S:T/24T/24
Dng 5: Tm thi im 0t vt c li x ( hay vn tc v).
A. Kin thc cn bn:
Thay x vo phng trnh 0cosx A t t
Thay v vo phng trnh 0sinv A t t
I IPhng php :
1.Phng php ng trn lng gic:a. Khi vtqua li x0th :Ta c th da vo mi lin h gia DH v CT Thng qua cc bc sau
* c 1 : V ng trn c bn knh R A (bin ) v trc Ox nm ngang
*c 2 : Xc nh v tr vtlc t 0 th 0
0
x ?
v ?
Xc nh v tr vtlc t (xt bit)
* c 3 : Xc nh gc qut MOM'?
* c 4 :0
T 360 2t ?
t T
0T T
2360
b.Ch : tnh thi gian vt i qua v tr x bit ln th nta c th tnh theo cngthc sau:
+Nu n l s l th : 11
2n
nt T t
vi t1l thi gian vt i t v tr x0(lc t=0) n v tr x ln th nht.
+Nu n l s chn th: 22
2n
nt T t
vi t2l thi gian vt i t v tr x0(lc t=0) n v tr x ln th hai.
c. Khi vtt vn tc v0th :
v0 -Asin(t +) sin(t +) 0v
A sinb t b k2
t ( b) k2
1
2
b k2t
d k2t
vi k N khib 0
b 0
v k N* khi
b 0
b 0
d.Sphn bthi gian chuyn ng ca vt trn quo dao ng(cho kt qunhanh hn)
- Dng s ny c thgii nhanh vthi gian chuyn ng, qung ng i c trong thi gian t,qung ng i ti a, ti thiu.
- C thp dng c cho dao ng in, dao ng in t.- Khi p dng cn c knng bin i thi gian cho t lin hvi chu kT. v ch chng i xng
nhau qua gc ta .
M, t = 0
M , tv < 0
x0 x
v < 0
v > 0
x0O
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2.Phng php i s:Xc nh thi im vt qua v tr v chiu bit
-Vit ccphng trnh x v v theo t :
)sin(
)cos(
tv
tAx
-Nu vt qua x0v i theo chiu dng th
0)sin(
)cos(0
tv
tAx
(1)
-Nu vt i qua x0v i theo chiu m th
0)sin(
)cos(0
tv
tAx
(2)
-Gii (1) hoc (2) ta tm c t theo k( vi 2...1,0,k )-Kt hp vi iu kin ca t ta s tm c gi tr k thch hp v tm c t.
C th: x0Acos(t +) cos(t +) 0x
Acosb t + b +k2
* t1b
+
k2
(s) vi k N khi b > 0(v < 0) vtqua x0theo chiu m
* t2b
+
k2
(s) vi k N* khi b < 0 (v > 0) vtqua x0theo chiu dng
kt hp vi iu kin cabai ton ta loi bt i mt nghim
B. Cc v d:
V d1:Mt vt dao ng iu ha vi phng trnh: 4cos 43
x t cm
. Tm nhng thi im vt qua v
tr cn bng?
Hng dn gii:Khi vt qua vtr cn bng:
10 4cos 4 0 cos 4 0 4 0,1,2,3,...
3 3 3 2 24 4
kx t t t k t k
V d 2:Mt vt dao ng iu ha theo phng trnh x 5cos 4 t (cm). Vt qua v tr cn bng theochiu dng vo nhng thi im no ? Khi ln vn tc bng bao nhiu ?
Hng dn gii:Khi vt qua v tr cn bng th x = 0
nn: 5cos 4 t 0 cos 4 t cos 4 t2 2
V vt qua v tr cn bng theo chiu dng nn v > 0
34 t k2 t 0,5k 2 8
vi k Z . Khi : maxv A 4 .5 20 cm/s V d3: Mt cht im dao ng iu ha theo phng trnh:x = 5cos 10 t (cm). Thi im cht im qua imM1c li x1= - 2,5 cm ln th nht l:
sA60
1. sB
15
1. sC
6
1. sD
60
11.
Hng dn gii1:
Th li x1= - 2,5 cm vo phng trnh dao ng ta c:2
110cos10cos55,2 nn tt
1
2
210 2 ;3
210 2 ;
3
t k k Z
t k k Z
=>1 1min
2 2min
2 1; 030 15 15
2 2; 3
30 15 15
kt k t
kt k t
Chon B
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Hng dn gii 2:Dng vng trn lng gic : V hnh vi gc quay 2/3 => Thi gian l T/3 =1/15s
V d 4:Mt vt dao ng iu ha vi phng trnh x 20cos 10 t2
(cm). Xc nh thi im u tin
vt qua v tr c li x = 5 cm theo chiu ngc vi chiu dng k t thi im t = 0. Hng dn gii:
Ta c: 120cos 10 t 5 cos 10 t cos 0,422 2 4
V v < 0 nn 10 t 0,42 k22 t 0,008 0,2k vi k Z .
V t > 0 nn vt qua v tr c li x = 5 cm ln u tin ng nghim dng nh nht trong h nghim ny l k = 1.Vy t = 0,192 s.
V d 5:Mt vt dao ng iu ha vi phng trnh x 4cos 10 t3
(cm). Xc nh thi im gn nht
vn tc ca vt bng 20 3 cm/s v tng k t lc t = 0.
Hng dn gii:Ta c: v x ' 40 sin 10 t
3
20 3 40 sin 10 t 20 3 40 cos 10 t3 6
3cos 10 t cos
6 2 6
V v tng nn:1
10 t k2 t 0,2k 6 6 30
vi k Z
V t > 0 nn thi im gn nht l 1
t s6
.
V d 6:Mt vt dao ng iu ha theo phng trnh x 8cos2 t (cm). K t t = 0 vt qua v tr cn bng lnth nht ti thi im ?
Hng dn gii:Ta c:1 1
0 8cos2 t cos2 t 0 2 t k t k 2 4 2
vi k N
V t > 0 nn k = 0, 1, 2, 3, ...
Vt qua v tr cn bng ln th nht ng vi k = 0 1
t s4
V d 7:Mt vt dao ng iu ha theo phng trnh x 4cos 4 t 6
( x tnh bng cm v t tnh bng s).
K t t = 0, vt qua v tr x = 2 cm ln th ba theo chiu dng vo thi im no ? Hng dn gii1:Vt qua v tr x = 2 cm theo chiu dng nn v > 0, ta c 2 iu kin:
12 4cos 4 t cos 4 t
x 2 6 6 24 t k2
v 0 6 324 sin 4 t 0 sin 4 t 0
6 6
1 1t k
8 2
vi k = 1, 2, 3, 4, ...
Vt qua v tr x = 2 cm ln th ba ng vi k = 3 1 1 1 3 11t .3 s8 2 8 2 8
Hng dn gii2:S dng mi lin h gia chuyn ng trn u vi dao ng iu ha.
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P
-A A x
A/2
M
O
N
Lc t = 0 vt v tr c li l 4 3x A cm2
ngvi
vt v tr M.Vt qua v tr x = 2 cm theo chiu dng tc l qua im PVt qua im P ln th ba ng vi gc qut l:
2.2 2 MOP . Vi MOP6 3 2
Vy, 3 114 2 42 2 2
Thi im vt qua v tr x = 2 cm ln th ba l: 11
112t s4 8
V d 8:Mt vt dao ng iu ha theo phng trnh x 10cos 2 t2
(cm). Tm thi im vt qua v tr
c li x = 5 cm ln th hai theo chiu dng.Hng dn gii:
Ta c: 15 10cos 2 t cos 2 t cos2 2 2 3
1t k
122 t k2
52 3t k
12
vi k Z v t > 0 k = 1, 2, 3, ...
V qua v tr x = 5 cm theo chiu dng nn v > 0
Khi , 20 sin 2 t 02
. tha mn iu kin v > 0, ta chn:5
t k12
Vt qua v tr x = 5 cm ln th hai nn k = 2 : Vy: 5 19t 2 s12 12
V d 9:Vt dao ng iu ha theo phng trnh x 5cos t (cm) s qua v tr cn bng ln th ba (k t lct = 0) vo thi im no ?
Hng dngii:
Ta c: 0 5cos t cos t 0 t k 2
1t k
2 vi k Z V t > 0 nn k = 0, 1, 2, 3, ...
Vt qua v tr cn bng ln th ba ng vi k = 2 Vy 1
t 2 2,5 s2
V d 10:Mt cht im dao ng iu ha theo phng trnh2
x 4cos t3
(x tnh bng cm v t tnh bng s).
K t t = 0, cht im qua v tr c li x 2 cm ln th 2011 ti thi im ?Hng dn gii:
Ta c: 2 2 1 22 4cos t cos t cos3 3 2 3
t 1 3k 2 2t k2t 1 3k 3 3
vi k Z
Vi k = 0 th vt qua v tr x = 2 cm ln th nht ti thi im t = 1 sVi k = 1 th vt qua v tr x = 2 cm ln th hai v ba ti thi im 2 s v 4 s
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P
-A A x
A/2
M
O
N
Vy vt qua v tr x = 2 cm ln th 2011 ng vi k =1005. Suy ra, t = 1 + 3.1005 = 3016 s.
V d 11:Mt vt dao ng iu ha theo phng trnh x 4cos 4 t6
( x tnh bng cm v t tnh bng s).
K t t = 0, vt qua v tr x = 2 cm ln th 2013 vo thi im l bao nhiu ?Hng dn gii:
Ta c: 12 4cos 4 t cos 4 t cos 4 t k2
6 6 2 3 6 3
1 14 t k2 t k
6 3 24 2
1 1t k4 t k2
8 26 3
Vt qua v tr x = 2 cm ln th 2013 ng vi k = 1006 nghim trn.
Vy 1 1 1 12073t .1006 503 s24 2 24 24
Cch 2: S dng mi lin h gia chuyn ng trn u vi dao ng iu ha.
Lc t = 0 vt v tr c li A 3 4 3x cm2 2 Mi chu k (1 vng) vt qua v tr x = 2 cm l 2 ln Qua v tr x = 2 cm ln th 2013 th vt phi quay 1006vngri tip tc i t M n N, tc gc qut l:
120731006.2 2012
6 6 6
Suy ra:
12073120736t s
4 24
V d 12:Mt vt dao ng iu ha theo phng trnh x 10cos 10 t2
(cm). Xc nh thi im vt
qua v tr x = 5 cm ln th 2014.Hng dn gii:
Ta c: 15 10cos 10 t cos 10 t cos2 2 2 3
1 110 t k2 t k
2 3 60 510 t k2
5 12 310 t k2 t k
2 3 60 5
vi k Z
V t > 0 nn khi vt qua v tr x = 5 cm ln th 2014 ng vi k = 1007
Vy 1 1 1 1007 12083t k 201,383 s60 5 60 5 60
V d 13:Mt vt dao ng iu ho vi phng trnh x = 4cos(4t +6
)cm. Thi im th 2014 vtqua v tr
x=2cm l( khng xt theo chiu):
A)4027
8s B)
1007
4s C)
1007
2s D)
12085
24s
Gii Cch 1:*
1
4 2 k N6 3 24 221
k N4 28 26 3
k
t k txk
tt k
O x
M1
M2
A
-A
M0
Hnh 13
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Vt qua ln th 2014(CHN) ng vi nghim DI2014
10072
k 1 4027
503,5 = s8 8
t -> p n A
Gii Cch 2:Vt qua x =2 l qua M1v M2. Vt quay 1 vng (1 chu k) qua x = 2 l 2 ln.Qua ln th 2014 th phi quay 1006 vng ri i t M0n M2.(Hnh 2 : gc M0OM2=3/2)
Gc qut :3 3 4027
1006.2 5032 8 8
t s
p n A
V d 14:Mt vt dao ng iu ha theo phng trnh: x = 2cos(2t -2
3
) cm. Tm thi im vt qua v tr c li
x = 3 v ang i theo chiu (-) ln th 20.
Hng dn:
Cch gii 1: Khi vt qua v tr c li x = 3 ; v < 0:
22 2
3 6t k
(do v < 0 nn ta loi nghim
22 2
3 6t k
):
50;1;2;...
12t k k
Vt qua li x = 3 v ang i theo chiu (-) ln th 20 ng vi k = 19: 205
19 19,4212
t s
Cch gii 2: Ti thi im t = 0 vt qua li :2
2 os( ) 1( )3
x c cm
v
24 sin( ) 2 3( / ) 03v cm s
(theo chiu +)
Vt qua li x = 3 v ang i theo chiu(-) ln th 1 vo thi im: 1 512 4 12 12T T T T t
Vt qua li x = 3 v ang i theo chiu (-) ln th 2 vo thi im: 2 1 1.t t T
Vt qua li x = 3 v ang i theo chiu (-) ln th 20 vo thiim:
t20= t1+ 19T= 5T/12 + 19T = 19,42(s)
C.BI TP TLUYN DNG 5
Cu 1.Mt vt dao ng iu ho c phng trnh x 8cos(2t) cm. Thi im th nht vt i qua v tr cn bngl :
A.1
4s. B.
1
2s C.
1
6s D. 1
3s
22 os(2 ) 3
3x c t
24 sin(2 ) 03
v t
2 3os(2 )
3 2c t
24 sin(2 ) 03
v t
t1
t2
3
2
A
0v -1 x
2-2 O
-A A3
2
A
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Cu 2: Cho mt vt dao ng iu ha c phng trnh chuyn ng
6
t210cosx (cm). Vt i qua v tr
cn bng ln u tin vo thi im:
A.1/3 (s) B.1/6(s) C.2/3(s) D.1/12(s)
Cu 3.Mt vt dao ng iu ha c phng trnh x 8cos10t(cm). Thi im vt i qua v tr x 4cm ln th
2013 k t thi im bt u dao ng l :
A.6037
30(s). B.
6370
30(s) C.
6730
30(s) D.
603,7
30(s)
Cu 4:(H 2011)Mt cht im dao ng iu ha theo phng trnh2
4cos3
x t
(x tnh bng cm; t tnh
bng s). K t t = 0, cht im i qua v tr c li x = -2 cm ln th 2011 ti thi im
A.3015 s. B.6030 s. C.3016 s. D.6031 s.
Cu 5. Mt vt dao ng iu ho vi phng trnh x = 4cos(4t +6
)cm. Thi im th 2011 vt qua v tr
x=2cm.
A.12061
24s B.
12049
24s C.
12025
24s D. p n khc
Cu 6: Mt vt dao ng iu ho vi phng trnh x = 4cos(4t +6
)cm. Thi im th 2013 vt qua v tr
x=2cm l( khng xt theo chiu):
A.12073
24s B.
12061
24s C.
24157
24s D. p n khc
Cu 7: Mt vt dao ng iu ha theo phng trnh x=10cos(10.t) (cm).Thi im vt i qua v tr N c li
x=5 cm ln th 2013 theo chiu dng l:
A.401,8s B.402,60s C.410,78s D.402,567s
Cu 8:Mt vt dao ng iu ha theo phng trnh x=10cos(10.t) (cm).Thi im vt i qua v tr N c li x=5
cm ln th 2009 theo chiu dng l:
A.401,8s B.408,1s C.410,8s D.401,77s
Cu 9: Mt dao ng iu ho vi x=8cos(2t-6
) cm. Thi im th 2014 vt qua v tr c vn tc v= - 8cm/s.
A. 1006,5s B.1005,5s C.2014 s D. 1007s
Cu 10. Mt cht im dao ng iu ho trn trc Ox c vn tc bng 0 ti hai thi im lin tip 1 1,75t s v2 2,5t s , tc trung bnh trong khong thi gian l 16 /cm s . To cht im ti thi im 0t l
A. -8 cm B. -4 cm C. 0 cm D.-3 cm
Cu 11: Mt vt dao ng c phng trnh l2
3cos(5 ) 1( )3
x t cm
. Trong giy u tin vt i qua v tr
c ta l x=1cm my ln?A. 2 ln .3 ln C.4 ln D.5 ln
Cu 12: Mt cht im dao ng iu ho vi phng trnh x = 4cos(2t + /2)cm. Thi gian t lc bt u dao
ng n lc i qua v tr x = 2cm theo chiu dng ca trc to ln th 1 l
A.0,917s. B.0,583s. C.0,833s. D.0,672s.
Cu 13.Mt cht im dao ng iu ho vi phng trnh li x = 2cos(t) cm. Vt qua v tr cn bng ln th
nht vo thi im
A. t = 0,5 (s). B. t = 1 (s). C. t = 2 (s). D. t = 0,25 (s).
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Cu 14.Mt vt dao ng iu ho vi phng trnh x = 4cos(0,5t - 5/6) cm. Vo thi im no sau y vt s
qua v tr x = 2 3 cm theo chiu m ca trc to .
A. t = 1 s. B. t = 4/3 s. C. t = 1/3 s. D. 2 s.
Cu 15.Mt vt dao ng iu ho vi phng trnh )5cos(4 tx (cm). Thi im u tin vt c vn tc bng
na ln ca vn tc cc i l:
A.3011 s. B.
307 s. C.
61 s. D.
301 s.
Cu 16.Vt nng trong con lc l xo dao ng iu ha vi phng trnh cos(4 )6
x A t cm
. Thi im cht
im i qua v tr c ng nng bng th nng ln 2014 v 2015 ln lt l bao nhiu?
p n & Hng dn chi tit:
Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8 Cu 9 Cu 10
A A A C A A A D A D
Cu 11 Cu 12 Cu 13 Cu 14 Cu 15 Cu 16 Cu 17 Cu 18 Cu 19 Cu 20
D B A D D
Cu 1. Gii:Chn A
Cch 1 :Vt qua VTC: x 0 2t /2 +k2 t 1
4+k vi k N
Thi im th nht ng vi k 0 t 1/4 (s)
Cch 2 :S dng mi lin h gia DH v CT.B1 V ng trn (hnh v1)
B2 Lc t 0 : x08cm ; v00 (Vti ngc chiu +t v tr bin dng)
B3 Vt i qua VTC x 0, v < 0
B4 Vt i qua VTC, ng vi vt chuyn ng trn u qua M0v M1. V 0, vt xut pht t M0nn thi
im th nht vt qua VTC ng vi vt qua M1.Khi bn knh qut 1 gc 2
t
0360
T 1
4s.
Cu 2.Chn :A Gii: t = 0 : 5 3 , 0x cm v ;2 1
23 3
t t s
Cu 3.Chn A Gii:
Cch 1:*
1 k10 t k2 t k N
3 30 5x 4
1 k10 t k2 t k N
3 30 5
Vt qua ln th 2013 (l) ng vi v tr M1: v 0, ta chn nghim trn
vi2013 1
k 10062
t
1
30+
1006
5
6037
30s. Chn : A
Cch 2 :Lc t 0 : x08cm, v00
Vt qua x 4cm l qua M1v M2.Vt quay 1 vng (1chu k) qua x 4cm l 2 ln. Qua ln th 2013 th phi quay 1006 vng ri i t M0n
M1.Gc qut 1 6037
1006.2 t (1006 ).0,2 s3 6 30
. Chn : A
AA
M
x
M
M2
O
Hnh 1
O x5 3
10-10
AA
M
x
M
M2
O
Hnh 3
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Cu 4. Chn : C Gii: t = 0 : x = 4cm , v < 0
V tr x = -2 cm th 1: 2 2
13 3
t t s
23T s
. Mt chu k qua x =-2cm : 2 ln. Ln th 2011 ng vi t = 1+1005x3 = 3016s
Cu 5. Gii Cch 1:*
14 2 k N
6 3 24 221
k N4 28 26 3
kt k t
xk
tt k
Vt qua ln th 2011(l) ng vi nghim trn2011 1
10052
k
1 12061
502,5 = s24 24
t -> Chn : A
Gii Cch 2:Vt qua x =2 l qua M1v M2. Vt quay 1 vng (1 chu k) qua x = 2 l 2 ln.Qua ln th 2011 th phi quay 1005 vng ri i t M0n M1.(Hnh 5)
Gc qut1 12061
1005.2 502,5
6 24 24
t s
Chn : A
Cu 6. Gii Cch 1:*
14 2 k N
6 3 24 22
1k N4 2
8 26 3
kt k t
xk
tt k
Vt qua ln th 2013(l) ng vi nghim trn2013 1
10062
k
1 12073
503 = s24 24
t -> p n A
Gii Cch 2:Vt qua x =2 l qua M1v M2. Vt quay 1 vng (1 chu k) qua x = 2 l 2 ln.Qua ln th 2013 th phi quay 1006 vng ri i t M0n M1.(Hnh 6 : gc M0OM1=/6)
Gc qut : 1 120731006.2 5036 24 24
t s
. p n A
Cu 7. Gii: Chu k T= 0,2s.lc u vt M0.
-V vng trn lng gic :
-Thi im vt qua v tr N c li x=5 cm ln th 1 theo chiu dng l:
Gc quay 5/3 ng 5T/6 = 5.0,2/6= 1/6s
-Thi gian vt i qua v tr N c li x=5 cm t ln th 2 n ln 2013 theo chiu dng l:
2012.T = 2012.0,2= 402,4 s.
-Vy Thi im vt i qua v tr N c li x=5 cm ln th 2009 theo chiu dng l:
402,4s+ 1/6s =12077/30 s=402,5666667 s .Vy chn D
Cu 8: T pt:x=10cos(10.t) (cm).=>= 0. Chu k : 2 2
0,210
T s
=> vt v tr nm ngang. V tr c li x = 5cm v theo chiu dnghp vi trc ngang gc -/3Khi vt quay 1 vng (tc l 1 chu k T) th n i qua v tr c li x = 5cm theo chiu dng 1 ln .
=> Vt quay 2008 vng tc l 2008T th n s qua v tr c li x = 5cm theo chiu dng 2008 ln.
Cn vng cui ln 2009 th n s quay c 1 gc 5/3 ng :5T/6 = 5.0,2/6= 1/6s
Ox
-2 4-4
O x
M1
M2
A-A
M0
Hnh 5
O x
M1
M2
A-A
M0
Hnh 6
Hnh 7
O x
M
N
10-10
M0
5
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Thi im vt i qua v tr N c li x=5 cm ln th 2009 theo chiu dng l:
-Vy Thi im vt i qua v tr N c li x=5 cm ln th 2009 theo chiu dng l:
t= 2008T +T/6 = 401,6+ 1/6 =12053/30 s=401,7666667 s .Vy chn D
Cu 9.Chn AGai:
Cch 1:Ta c v = -16sin(2t-6
) = -8
12 26 6 6
5 1
2 26 6 2
t k t k
k N
t k t k
Thi im th 2012 ng vi nghim2014
1 10062
k 1
1006 1006,52
t s .Vy chn A
Cch 2:Ta c 2 2( ) 4 3v
x A cm
.V v < 0 nn vt qua M1v M2; Qua ln th 2014 th phi quay 1006
vng ri i t M0n M2. Gc qut = 1006.2+ t = 1006,5 s . (Hnh 9)
Cu 10.Gii: Gi s ti thi im t0 = 0;, t1v t2cht im cc v trM0; M1v M2;t thi im t1n t2cht im C theo chiu dng.
Cht im c vn tc bng 0 ti cc v tr bin .Chu k T = 2(t2t1) = 1,5 (s)vtb= 16cm/s. Suy ra M1M2 = 2A = vtb(t2t1) = 12cm
Do A = 6 cm. T t0= 0 n t1: t1= 1,5s + 0,25s = T + T6
1
V vy khi cht im M0, cht im C theo chiu m, n v tr bin m ,trong t=T/6 i c qung ng A/2. Do vy ta cht im t thi im t = 0l x0= -A/2 = - 3 cm. Chn D
Cu 11.Gii:Vt dao ng ha quanh v tr x=1cm
Ta c:2
25,22
55
2
1 TTTt
T
t
; thi im t=0 )1(
0
2
1
v
cmx
Trong 2 chu k vt qua v tr x=1cm c 4 ln( mi chu k qua 2 ln)Trong na chu k tip theo vt qua x=1cm thm 1 ln na.Chn D
Cu 12.Gii: Chn Bt = 0 : x = 0 , v < 0
x = 2cm , v > 07 7
2
6 12
t t s
Cu 13.Chn A Cu 14.Chn D Cu 15.Chn D.
Cu 16.Ta c thgii bi ton ny theo 2 cch nh sau:
Cch gii theo vng trn:
+ ti t=0 th 0
0
3
3
0
Ax
v
tng ng
vi im M0 trn vng trn.
( vi T=0,5s)
M0 M2M1
O x2
4
-4
O
A A
2013
2M
x
0M
2
2
A
2
2
A
2014
2M
2015
2M 2016
2M
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+ khi w wd t th 2 21 1 2
2 22 2 2
t d tW W W W kA kx x A
Do c 2 ta nn trong mt chu ksc 4 ln ng nng bng thnng. V vy phng php gii vng trn tastch sln bi thnh slin k, nhhn n nhng chia ht cho 4 ( bi ca 4) vi mc ch tm schu kdaong u tin v lng d cn li ri tm nt khong thi gian tng ng. Cth nh sau:
+ i vi ln th2014ta vit tch thnh 2012 (v 2012 l schia ht cho 4, lin kv nhhn 2014) thi
im ng nng bng thnng ln th2014 c tnh l 2014 22012
4t T t ,trong t2l khong thi gian
dch chuyn trn cung M0M2. Ta c: 2 0 27
6 8 24M M
T Tt t T .Vy: 2014
2012 7 12079 12079
4 24 24 48
T Tt T s
+ i vi ln th2015th ta li vit tch thnh 2012+3 thi im ng nng bng thnng ln th2015 c
tnh l 2015 32012
4t T t , trong t3l khong thi gian dch chuyn trn cung M0M2(2015).
Ta C 3 0 2(2015) 136 4 8 24
M M T T Tt t T Nn 2015 2012 13 12085 120854 24 24 48Tt T T s
Cch gii theo cng thc tnh nhanh:
V tr : 2
2t dW W x A . Do 2 v tr ny i xng nhau qua VTC nn ta c th quan nim bi ton ny
l tm thi im ln th 2014 v 2015 vt cch VTC mt khong 2
2L A .
+ i vi ln th 2014 th :
2014
5034 T T d 2 nn ta c: 2014 2 503t t T Theo hnh v th:
D dng c 27
6 8 24
T Tt T =>
2014
7 12079 12079503
24 24 48
T Tt T s
+ i vi ln th 2015 th :2015
5034
T T d 3 nn ta c: 2015 3 503t t T
D dng c 313
6 4 8 24
T T Tt T => 2015
13 12085 12085503
24 24 48
Tt T T s
ln 3
ln 2( ln 1)
A O A
x
2
2
A2
2
A 3
2
A
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Dng 6: Tm li ca vt sau khong thi gian t .
A. Kin thc cn bn:
Ti thi im1t vt c li 1x Tm li 2x ca vt sau khong thi gian t .
Cch 1:
Ti
1 1
1 1 2 2 2 1
1 1
cos: ? 1 ; : cos cos 2
sin
x A tt t t x A t A t t
v A t
Th(1) vo (2) 2x . Xc nh vtr 1x ca vt trn trc ox. T 2t nT x
Cch 2:Phng php nhanh:Tnh lch pha gia x1v x2 : = .t
*Xt lch pha: +Nu (c bit)
x2v x1cng pha x2= x1
x2v x1ngc pha x2= - x1
x2v x1vung pha 2 2 21 2x x A .
+Nu bt k: Bm my tnh Fx 570ES vi ch : 4SHIFT MODE : n vgc l Rad.
*Bm nhp my tnh: 12 cos cos( )x
x A SHIFTA
.
*Quy c du trc shift: du (+) nu x1
du (-) nu x1
Nu khng ni ang tng hay ang gim, ta ly du +
Cch 3: Nhcc trng hp c bit:
B. Cc v d:
V d1:Mt vt dao ng iu ha vi phng trnh: 6cos 24
x t cm
. Ti thi im1
t vt c li
1 4x cm , tm li ca vt thi im: 2 1 4,5t t s
Hng dn :
Cch 1: 1 16cos 2 44
x t cm
2 2 1 1 16cos 2 6cos 2 4,5 6cos 2 9 6cos 2 44 4 4 4
x t t t t cm
4
T
12
T
6
T
A A
2
2
A 0 2
A 3
2
A
8
T
8
T
6
T
12
T
x
2
T
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Cch 2:Chu k 2 12
1 4,5 4 0,5 4T s t s T T x x cm
.
V d2:Mt vt nhdao ng iu ha vi 6cos 2x t cm . Ti thi im t1no , li ca vt l
-3cm. Ti thi im t2 = t1+0.5s,vn tc ca vt c gi tr:
A: 6 3 /cm s B:-3m/s C:3cm/s D: 3 3 /cm s
Gii: Chu k 2 1 1 2 12
1 0,5 0,5 3T s t t s t T x x cm
.
2 2 2 22 6 3 6 3 /v A x cm s .Chn A
V d3:Mt dao ng iu ha x = 10cos(4t3/8) cm. Khi t = t1th x = x1= -6cm v ang tng. Hi,khi t = t1+ 0,125s th x = x2= ?
Gii:Cch 1:Dng lch pha.Tnh = 4.0,125 = /2 (rad) x1v x2vung pha.
2 2 2 2 2
1 2 2x x A x 10 ( 6) 8cm . M x1nn x2= 8cm.
Cch 2:Bm my tnh Fx570Es: 1 610 cos shift cos10 2
= 8 x2 = 8cm.
V d4:Mt vt dao ng iu ha x = 5cos(4t/6) cm. Khi t = t1th x = 3cm v ang tng.
Hi, khi t = t1+1
12s th x2= ?
Gii 1:Dng lch pha: = .t = 4.1
12=
3
khng ng cho 3 trng hp c bit.
Gii 2:Bm my tnh Fx570Es:1 3
5 cos shift cos 4,9645 3
2 4,964x cm
C.BI TP TLUYN DNG 6
Cu 1:Mt vt nhdao ng iu ha vi