Phng Php Gii Ton Vt L 12Phn 1.DAO NG C HCCh 1 .DAO NG IU HA.CON
LC L XOA. KIN THC C BN. 1. Phng trnh dao ng c dng : . ( ) x Acos t
+ Trong : + A l bin dao ng.+ l vn tc gc, n v (rad/s).+ l pha ban u
( l pha thi im t = 0),n v (rad).+ x l li dao ng thi im t.+ (.t +) l
pha dao ng ( l pha thi im t).2. Vn tc trong dao ng iu ho.'. .sin( )
cos( )2v x A t A t + + + ; + v bin thin cng tn s, sm pha/ 2 so vi
x.+ vmax = A x = 0 ( Ti VTCB )+ vmin = 0 x =t A ( Ti hai bin )3.
Gia tc trong dao ng iu ho. ' " 2 2. . ( . ) . a v x A cos t x + + a
bin thin iu ha cng tn s, ngc pha so vi x+ amax = 2A x =t A ( Ti hai
bin )+ amin = 0 x = 0 ( Ti VTCB )+ ar lun c hng v VTCB. A lun ngc
du vi x 4. th: ) (t x, ) (t v, ) (t a:- Gi s vt dao ng iu ha c phng
trnh l: ) cos( + t A x.- n gin, ta chn = 0, ta c:t A x cos .t A x a
t A t A x v cos ' ' )2cos( sin '2 .- Vn tc v v gia tc a cng bin
thin iu ha vi cng tn s gc . Kho st ton hc ta v c th: - th cng cho
thy sau mi chu k dao ng th ta x, vn tc v v gia tc a lp li gi tr c.
Mt s gi tr c bit ca x, v, a nh sau:t 0 T/4 T/2 3T/4 Tx A 0-A 0 Av 0
-A 0 A 0a A2 0 A2 0 A2
Nguyn nh Khang1Phng Php Gii Ton Vt L 12 5. Cc h thc lin h c lp
thi gian gia x , v, a:2 2 22 2 2 22 2 2 2; 1; ..v x vA x v A xA A +
+ t +2 224 2 a vA6. Chu k dao ng: 2. 12. . .mTk f 7. Tn s dao ng :
1 1. .2. 2.kfT m 8. Lc trong dao ng iu ho :+ Lc n hi :. . . os( . )
.dhF k l x k l Ac t t t +
+ Lc phc hi : 2 2. . . . . . os( . ).phF k x m x m Ac t +
9. Nng lng trong dao ng iu ho : E = E + EtTrong : + E = 2 2 2 2
2 21 1 1 1. . . . . . os ( . ). os (2 . 2 )2 2 4 4mv m A c t kA kA
c t + + +L ng nng ca vt dao ng.E bin thin tun hon vi tn s = 2+ Et =
2 2 2 2 21 1 1 1. . . . .sin ( . ) .cos (2 . 2 ).2 2 4 4k x k A t
kA kA t + +L th nng ca vt dao ng ( Th nng n hi ).Et bin thin tun
hon vi tn s = 2
2 2 21 1. . . . .2 2d tE E E m A k A const + . 10. Cc loi dao ng
: + Dao ng tun hon. + Dao ng iu ho.+ Dao ng t do. + Dao ng tt dn.+
Dao ng cng bc. + S t dao ng.Nguyn nh Khang2Phng Php Gii Ton Vt L
12B. BI TP DNG 1.XC NH CC C IM TRONG DAO NG IU HOI.Phng php.+ Nu u
bi cho phng trnh dao ng ca mt vt di dng c bn :. os( . ), x Ac t +
th ta ch cn a ra cc i lng cn tm nh : A, x, ,.+ Nu u bi cho phng
trnh dao ng ca mt vt di dng khng c bn th ta phi p dng cc php bin i
lng gic hoc php i bin s ( hoc c hai) a phng trnh v dng c bn ri tin
hnh lm nh trng hp trn.II. Bi Tp.B i 1 . Cho cc phng trnh dao ngiu
ho nh sau. Xc nh A, , , f ca cc dao ng iu ho ?a) 5. os(4. . )6x c t
+(cm). b) 5. os(2. . )4x c t +(cm).c)5. os( . ) x c t (cm).d)
10.sin(5. . )3x t +(cm).L i Gii: a)5. os(4. . )6x c t +(cm).5( );
4. ( / ); ( );6A cm Rad s Rad 2. 2. 1 10, 5( ); 2( )4. 0, 5T s f
HzT b)5.5. os(2. . ) 5. os(2. . ) 5. os(2. . ).4 4 4x c t c t c t +
+ + + (cm). 5.5( ); 2. ( / ); ( )4A cm rad s Rad 2. 11( ); 1( ). T
s f HzT c)5. os( . )( ) 5. os( . )( ) x c t cm c t cm + 2.5( ); ( /
); ( ); 2( ); 0, 5( ). A cm Rad s Rad T s f Hz d) 10.sin(5. . ) 10.
os(5. . ) 10. os(5. . )3 3 2 6x t cm c t cm c t cm + + .2. 110( );
5. ( / ); ( ); 0.4( ); 2, 5( )6 5. 0, 4A cm Rad s Rad T s f Hz .B i
2 . Cho cc chuyn ng c m t bi cc phng trnh sau:a)5. ( . ) 1 x cos t
+(cm)b) 22.sin (2. . )6x 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. Xc
nh bin , tn s, pha ban u, v v tr cn bng ca cc dao ng . Li Giia)5. (
. ) 1 x cos t + 1 5. ( . ) x cos t . t x-1 = X. ta c5. os( . ) X c
t l mt dao ng iu hoVi 5( ); 0, 5( ); 0( )2. 2.A cm f Hz Rad VTCB ca
dao ng l : 0 1 0 1( ). X x x cm b)22.sin (2. . ) 1 (4. . )6 3x t
cos t + + t X = x-1 os(4. . ) os(4 )6 3X c t c t + l mt dao ng iu
ho.Nguyn nh Khang3Phng Php Gii Ton Vt L 12Vi4.1( ); 2( ); ( )2. 2.
3A cm f s Rad c) 3.sin(4. . ) 3. (4. . ) 3.2sin(4. ). ( ) 3.
2.sin(4. . )( ) 3 2 os(4. . )( )4 4 4 4x t cos t t cos x t cm c t
cm + + + l mt dao ng iu ho. Vi 4.3. 2( ); 2( ); ( )2. 4A cm f s Rad
B i 3. Hai dao ng iu ho cng phng , cng tn s, c cc phng trnh dao ng
l: 13. os( . )4x c t (cm) v 24. os( . )4x c t +(cm) . Bin ca dao ng
tng hp hai dao ng trn l:A.5 cm. B.7 cm. C. 1 cm.D.12 cm. B i 4 .
Hai dao ng cng phng , cng tn s : 12 . os( . )3x a c t + (cm) v 2.
os( . ) x a c t +(cm) . Hy vit phng trnh tng hp ca hai phng trnh
thnh phn trn? A.. 2. os( . )2x a c t +(cm). B. . 3. os( . )2x a c t
+(cm).C.3.. os( . )2 4ax c t +(cm).D.2.. os( . )4 6ax c t +(cm).
DNG 2.XC NH LI , VN TC, GIA TC, LC PHC HI MTTHI IM HAY NG VI PHA
CHOI. Phng php.+ Mun xc nh x, v, a, Fph mt thi im hay ng vi pha d
cho ta ch cn thay t hay pha cho vo cc cng thc :. ( . ) x Acos t
+hoc.sin( . ) x A t +;. .sin( . ) v A t +hoc. . ( . ) v A cos t +
2. . ( . ) a A cos t + hoc2. .sin( . ) a A t +v .phF k x
.+Nuxcnhcli x, tacthxcnhgiatc, lcphchitheobiuthcnhsau: 2. a x v2. .
.phF k x m x + Ch : - Khi 0; 0;phv a F o f f f : Vn tc, gia tc, lc
phc hi cng chiu vi chiu dng trc to . - Khi 0; 0; 0phv a F p p p: Vn
tc , gia tc, lc phc hi ngc chiu vi chiu dng trc to . II. Bi Tp.B i
1 . Mt cht im c khi lng m = 100g dao ng iu ho theo phng trnh :5.
os(2. . )6x c t +(cm) . Ly 210. Xc nh li , vn tc, gia tc, lc phc hi
trong cc trng hp sau :a) thi im t = 5(s).b) Khi pha dao ng l
1200.Li GiiT phng trnh 5. os(2. . )6x c t +(cm)5( ); 2. ( / ) A cm
Rad s Vy2 2. 0,1.4. 4( / ). k m N m Nguyn nh Khang4Phng Php Gii Ton
Vt L 12Ta c '. . ( . ) 5.2. . (2. . ) 10. . (2. . )6 6v x A cos t
cos t cos t + + +a) Thay t= 5(s) vo phng trnh ca x, v ta c :
5.sin(2. .5 ) 5.sin( ) 2, 5( ).6 6x cm +
310. . (2. .5 ) 10. . ( ) 10. . 5. 306 6 2v cos cos +
(cm/s).
2 22 2. 4. .2, 5 100( ) 1( )cm ma xs s .Du chng t gia tc ngc
chiu vi chiu dng trc to . 2. 4.2, 5.10 0,1( ).phF k x N Du chng t
Lc phc hi ngc chiu vi chiu dng trc to . b) Khi pha dao ng l 1200
thay vo ta c :- Li : 05.sin120 2, 5. 3 x (cm).- Vn tc :010. . 120
5. v cos (cm/s).- Gia tc : 2 2. 4. .2, 5. 3 3 a x (cm/s2).- Lc phc
hi :. 4.2, 5. 3 0,1. 3phF k x (N).B i 2 . To ca mt vt bin thin theo
thi gian theo nh lut : 4. (4. . ) x cos t (cm). Tnh tn s dao ng ,
li v vn tc ca vt sau khi n bt u dao ng c 5 (s).Li GiiT phng trnh4.
(4. . ) x cos t (cm) Ta c :4 ; 4. ( / ) 2( )2.A cm Rad s f Hz .- Li
ca vt sau khi dao ng c 5(s) l : 4. (4. .5) 4 x cos (cm).- Vn tc ca
vt sau khi dao ng c 5(s) l : '4. .4.sin(4. .5) 0 v x B i 3 .Phng
trnh ca mt vt dao ng iu ho c dng : 6.sin(100. . ) x t +.Cc n v c s
dng l centimet v giy.a) Xc nh bin , tn s, vn tc gc, chu k ca dao
ng.b) Tnh li v vn tc ca dao ng khi pha dao ng l -300.B i 4.Mt vt
dao ng iu ho theo phng trnh :4.sin(10. . )4x t +(cm).a) Tm chiu di
ca qu o, chu k, tn s.b)Vo thi im t = 0 , vt ang u v ang di chuyn
theo chiu no? Vn tc bng bao nhiu?DNG 3. CT GHP L XOI. Phng php.B i
ton: Mt l xo c chiu di t nhin l0 , cng l k0 , c ct ra thnh hai l xo
c chiu di v cng tng ng l : l1, k1 v l2, k2. Ghp hai l xo vi nhau.
Tm cng ca h l xo c ghp.Nguyn nh Khang5mk1,l1k2,l2Phng Php Gii Ton
Vt L 12L i gii: + Trng hp 1 : Ghp ni tip hai l xo (l1 , k1 ) v ( l2
,k2).
1 21 2dh dhF F Fl l l + Ta c 1 1 1 2 2 2. ; . ; .dh dhF k l F k
l F k l .1 21 21 2; ; .dh dhF F Fl l lk k k Vy ta c : 1 21 2 1 21 1
1dh dhF F Fk k k k k k + + (1)+ Trng hp 2 : Ghp song song hai l xo
(l1 , k1 ) v ( l2 ,k2).
1 21 2dh dhF F Fl l l +
1 1 2 2 1 2. . . k l k l k l k k k + + (2) Ch : cng ca vt n hi c
xc nh theo biu thc :.Sk El(3)Trong : + E l sut Yng, n v : Pa, 2 2;1
1N NPam m .+ S l tit din ngang ca vt n hi, n v : m2.+ l l chiu di
ban u ca vt n hi, n v : m.T (3) ta c :k0.l0 = k1.l1 = k2.l2 = Const
= E.S. II. Bi Tp.B i 1 . Mt vt khi lng m treo vo l xo c cng k1 =
30(N/m) th dao ng vi chu k T1 = 0,4(s) .Nu mc vt m trn vo l xo c
cng k2 = 60(N/m) th n dao ng vi chu k T2 = 0,3(s). Tm chu k dao ng
ca m khi mc m vo h l xo trong hai trng hp:a) Hai l xo mc ni
tip.b)Hai l xo mc song song. Bi 2. Hai l xo L1,L2 c cng chiu di t
nhin. khi treo mt vt c khi lng m=200g bng l xo L1 th n dao ng vi
chu k T1 = 0,3(s); khi treo vt m bng l xo L2 th n dao ng vi chu k
T2 =0,4(s).1.Ni hai l xo trn vi nhau thnh mt l xo di gp i ri treo
vt m trn vo th vt m s dao ng vi chu k bao nhiu? Mun chu k dao ng ca
vt '1 21( )2T T T +th phi tng hay gim khi lng m bao nhiu?2. Ni hai
l xo vi nhau bng c hai u c mt l xo c cng di ri treo vt m trn th chu
k dao ng l bng bao nhiu? Mun chu k dao ng ca vt l 0,3(s) th phi tng
hay gim khi lng vt m bao nhiu?B i 3 . Mt lxo OA=l0=40cm, cng k0 =
100(N/m). M l mt im treo trn l xo vi OM = l0/4.1. Treo vo u A mt vt
c khi lng m = 1kg lm n dn ra, cc im A v M n v tr A v M .Tnh OA v OM
.Ly g = 10 (m/s2).2. Ct l xo ti M thnh hai l xo . Tnh cng tng ng ca
mi on l xo.3. Cn phi treo vt m cu 1 vo im no n dao ng vi chu k T =
. 210s.B i 4 . Khi gn qu nng m1 vo l xo , n dao ng vi chu k T1 =
1,2s. Khi gn qu nng m2 vo l xo , n dao ng vi chu k T2 = 1,6s. Hi
sau khi gn ng thi c hai vt nng m1 v m2 vo l xo th chng dao ng vi
chu k bng bao nhiu?DNG 4.VIT PHNG TRNH DAO NG IU HOI. Phng php.Phng
trnh dao ng c dng : . ( . ) x Acos t +hoc.sin( . ) x A t +.Nguyn nh
Khang6k1,l1mPhng Php Gii Ton Vt L 121. Tm bin dao ng A:Da vo mt
trong cc biu thc sau: + 22 2 2 2 221. ; . ; . . . ; . . ;2max max
maxvv A a A F m A k A E k A A x + (1) + Nu bit chiu di ca qu o l l
th 2lA . + Nu bit qung ng i c trong mt chu k l s th 4sA .Ch : A
> 0. 2. Tm vn tc gc :Da vo mt trong cc biu thc sau : + 2.2. .kfT
m . + T (1) ta cng c th tm c nu bit cc i lng cn li. Ch : -Trong thi
gian t vt thc hin n dao ng, chu k ca dao ng l : tTn- > 0 ; n v :
Rad/s 3. Tm pha ban u : Da vo iu kin ban u ( t = 0 ). Gi tr ca pha
ban u () phi tho mn 2 phng trnh :00. os. .sinx Acv A ' )
Ch : Mt s trng hp c bit :+ Vt qua VTCB : x0 = 0.+ Vt v tr bin :
x0 = +A hoc x0 = - A.+ Bung tay ( th nh ), khng vn tc ban u : v0 =
0. II. Bi Tp.B i 1 . Mt con lc l xo dao ng vi bin A = 5cm, chu k T
= 0,5s. Vit phng trnh dao ng ca con lc trong cc trng hp: a) t = 0 ,
vt qua VTCB theo chiu dng.b) t = 0 , vt cch VTCB 5cm, theo chiu
dng.c) t = 0 , vt cch VTCB 2,5cm, ang chuyn ng theo chiu dng. Li
GiiPhng trnh dao ng c dng : .sin( . ) x A t +.Phng trnh vn tc c dng
: '. . ( . ) v x A cos t + .Vn tc gc : 2. 2.4 ( / )0, 5Rad sT .a) t
= 0; 00.sin. .x Av A cos 00 5.sin5.4. . 0 v cos f 0 . Vy 5.sin(4. .
) x t (cm).b) t = 0; 00.sin. .x Av A cos 05 5.sin5.4. . 0 v cos f(
)2rad . Vy 5.sin(4. . )2x t +(cm).c) t = 0; 00.sin. .x Av A cos 02,
5 5.sin5.4. . 0 v cos f( )6rad .Vy 5.sin(4. . )6x t +(cm).B i 2 .
Mt con lc l xo dao ng vi chu k T = 1(s). Lc t = 2,5(s), vt quav tr
c li 5. 2 x (cm) vi vn tc 10. . 2 v (cm/s). Vit phng trnh dao ng ca
con lc.Nguyn nh Khang7Phng Php Gii Ton Vt L 12 Li GiiPhng trnh dao
ng c dng : .sin( . ) x A t +.Phng trnh vn tc c dng : '. . ( . ) v x
A cos t + .Vn tc gc : 2. 2.2 ( / )1Rad sT .ADCT :22 22vA x +2 22 22
2( 10. . 2)( 5. 2)(2. )vA x + + = 10 (cm). iu kin ban u : t =
2,5(s); .sin. .x Av A cos 5. 2 .sin10. . 2 .2. .AA cos tan 1 (
)4rad . Vy 10.sin(2. . )4x t +(cm). Bi 3. Mt vt c khi lng m = 100g
c treo vo u di ca mt l xo c cng k = 100(N/m). u trn ca l xo gn vo
mt im c nh. Ban u vt c gi sao cho l xo khng b bin dng. Bung tay
khng vn tc ban ucho vt dao ng. Vit phng trnh da ng ca vt. Ly g = 10
(m/s2); 210 . Li GiiPhng trnh dao ng c dng : .sin( . ) x A t +.
10010.0,1km (Rad/s).Ti VTCB l xo dn ra mt on l :2. 0,1.1010 ( ) 1
1100m gl m cm A l cmk .iu kin ban u t = 0 , gi l xo sao cho n khng
bin dng tc x0 = -l . Ta c t = 0 ;001 .sin. . 0x l Av A cos f (
)2rad . Vy sin(10. . )2x t (cm).B i 4 . Mt vt dao ng iu ho dc theo
trc Ox. Lc vt qua v tr c li 2 x (cm) th c vn tc . 2 v (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. Li GiiPhng 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 Acos v A a Acos . Ly a chia cho x ta c : ( /
) rad s .Ly v chia cho a ta c : 3.tan 1 ( )4rad (v cos < 0 )2 A
cm . Vy : 3.2.sin( . )4x t +(cm).B i 5.Mt con lc l xo l tng t nm
ngang, t VTCB ko l xo dn 6 cm . Lc t = 0 bung nh , sau 512su tin ,
vt i c qung ng 21 cm. Phng trnh dao ng ca vt l : A.6.sin(20. . )2x
t +(cm)B. 6.sin(20. . )2x t (cm)Nguyn nh Khang8Phng Php Gii Ton Vt
L 12 C.6.sin(4. . )2x t +(cm) D. 6.sin(40. . )2x t +(cm)B i 6 .Mt
con lc l xo treo thng ng gm mt vt m = 100g, l xo c cng k =
100(N/m). Ko vt ra khi VTCBmt on x= 2cm v truyn vn tc62,8. 3 v
(cm/s) theo phng l xo .Chn t = 0 lc vt bt u dao ng ( ly 2210; 10mgs
) th phng trnh dao ng ca vt l:A.4.sin(10. . )3x t +(cm)B.4.sin(10.
. )6x t +(cm)C. 5.4.sin(10. . )6x t +(cm)D.4.sin(10. . )3x t (cm)B
i 7 . Mt qu cu khi lng m = 100g treo vo l xo c chiu di t nhin l0 =
20cm, cng k = 25 (N/m).a) Tnh chiu di ca l xo to v tr cn bng. Ly g
= 10 (m/s2).b) Ko qu cu xung di, cch v tr cn bng mt on 6cm ri bung
nhra cho n dao ng. Tm chu k dao ng, tn s . Ly 210 .c) Vit phng trnh
dao ng ca qu cu chn gc thi gian l lc bung vt; gc to ti v tr cn bng,
chiu dng hng xung.B i 8 . Mt qu cu khi lng m = 500g c treo vo l xo
c chiu di t nhin l0 = 40cm.a) Tm chiu di ca l xo ti v tr cn bng,
bit rng l xo trn khi treo vt m0 = 100g, l xo dn thm 1cm. Ly g = 10
(m/s2). Tnh cng ca l xo.b) Ko qu cu xung di cch v tr cn bng 8cm ri
bung nh cho dao ng. Vit phng trnh dao ng (Chn gc thi gian l lc th
vt, chiu dng hng xung).B i 9 . Vt c khi lng m treo vo l xo c cng k
= 5000(N/m). Ko vt ra khi v tr cn bng mt on 3cm ri truyn vn tc
200cm/s theo phng thng ng th vt dao ng vi chu k 25T s .a) Tnh khi
lng m ca vt.b) Vit phng trnh chuyn ng ca vt . Chn gc thi gian l lc
vt qua v tr c li x = -2,5cm theo chiu dng.B i 10: Cho con lc l xo
dao ng iu ho theo phng thng ng vt nng c khi lng m = 400g, l xo c
cng k, c nng ton phn E = 25mJ. Ti thi im t = 0, ko vt xung di VTCB
l xo dn 2,6cm ng thi truyn cho vt vn tc 25cm/s hng ln ngc chiu dng
Ox (g = 10m/s2). Vit phng trnh dao ng?DNG 5. CHNG MINH MT VT DAO NG
IU HOI. Phng php. 1. Phng php ng lc hc.Nguyn nh Khang9mPhng Php Gii
Ton Vt L 12 + Chn HQC sao cho vic gii bi ton l n gin nht.( Thng chn
l TT Ox, O trng vi VTCB ca vt, chiu dng trng vi chiu chuyn ng). +
Xt vt VTCB :1 20 ... 0 hlnF F F F + + + ur uur uur uurchiu ln HQC
thu c phng trinh v hng: 1 2 3... 0nF F F F t t t + (1) + Xt vt thi
im t, c li l x :p dng nh lut 2 Newton, ta c: 1 2. ... .hl nF ma F F
F ma + + + uur r uur uur uur rchiu ln HQC thu c phng trinh v
hng:
1 2... .nF F F ma t t t (2)Thay (1) vo (2) ta c dng : " 2. 0 x x
+ . Phng trnh ny c nghim dng: . ( . ) x Acos t +hoc.sin( . ) x A t
+ t dao ng iu ho, vi tn s gc l .2. Ph ng php nng lng. + Chn mt phng
lm mc tnh th nng, sao cho vic gii bi ton l n gin nht. + C nng ca vt
dao ng l :E = E + Et 2 2 21 1 1. . . . . .2 2 2k A mv k x +(3) + Ly
o hm hai v theo thi gian t , ta c : ' ' ' '1 10 . .2. . . .2. . 0 .
. . .2 2m v v k x x mv v k x x + + .Mt khc ta c : x = v ; v = a =
x, thay ln ta c : 0 = m.v.a + k.x.v" "0 . . . 0km x k x x xm + + .
t 2km . Vy ta c :" 2. 0 x x + Phng trnh ny c nghim dng: . ( . ) x
Acos t +hoc.sin( . ) x A t +
Vt dao ng iu ho, vi tn s gc l . pcm. DNG 6.TM CHIU DI CA L XO
TRONG QU TRNH DAO NG.NNG LNG TRONG DAO NG IU HOI. Phng php. 1. Chiu
di: + Nu con lc l xo t nm ngang : lmax = l0 + A; lmin = l0 -A.+ Nu
con lc l xo t thng ng : 0 maxl l l A + + ;min 0l l l A + . 2. Nng
lng :+ ng nng ca vt trong dao ng iu ho
2 2 2 21 1. . . . . . ( . )2 2dE mv m A cos t +hoc 2 2 2 21 1. .
. . . .sin ( . )2 2dE mv m A t + + Th nng ca vt trong dao ng iu ho
: 2 2 2 21 1. . . . . .sin ( . )2 2tE k x m A t +hoc 2 2 2 21 1. .
. . . . ( . )2 2tE k x m A cos t + + C nng ca vt trong dao ng iu
ho: 2 2 21 1. . . . .2 2d tE E E k A m A Const + .II. Bi Tp.B i 1 .
Mt vt khi lng m = 500g treo vo l xo th dao ng vi tn s f= 4(Hz). a)
Tm cng ca l xo, ly 210. Nguyn nh Khang10Phng Php Gii Ton Vt L 12 b)
Bit l xo c chiu di t nhin l0 = 20cm v dao ng vi bin 4cm. Tnh chiu
di nh nht v ln nht ca l xo trong qu trnh dao ng. Ly g = 10(m/s2).
c) Thay vt m bng m = 750g th h dao ng vi tn s bao nhiu?B i 2 . Mt
qu cu khi lng m =1 kg treo vo mt l xo c cng k = 400(N/m). Qu cu dao
ng iu hovi c nng E = 0,5(J) ( theo phng thng ng ).a) Tnh chu k v
bin ca dao ng.b) Tnh chiu di cc tiu v cc i ca l xo trong qu trnh
dao ng. Bit l0 = 30cm.c. Tnh vn tc ca qu cu thi im m chiu di ca l
xo l 35cm. Ly g=10(m/s2).B i 3 . Mt qu cu khi lng m = 500g gn vo mt
l xo dao ng iu ho vi bin 4cm. cng ca l xo l 100(N/m).a) Tnh c nng
ca qu cu dao ng.b) Tm li v vn tc ca qu cu ti mt im, bit rng ni , ng
nng ca qu cu bng th nng.c) Tnh vn tc cc i ca qu cu.B i 4 . Mt vt c
khi lng m = 500g treo vo mt l xo c cng k = 50(N/m). Ngi ta ko vt ra
khi v tr cn bng mt on 2(cm) ri truyn cho n mt vn tc ban u v0 =
20(cm/s) dc theo phng ca l xo.a) Tnh nng lng dao ng.b) Tnh bin dao
ng.c) Vn tc ln nht m vt c c trong qu trnh dao ng.B i 5 . Mt con lc
l xo c khi lng m= 50g dao ng iu ho theo phng trnh : 10.sin(10. .
)2x t +(cm) .a) Tm bin , tn s gc, tn s, pha ban u ca dao ng.b) Tm
nng lng v cng ca l xo. Bi 6. Mt con lc l xo dao ng iu ho bit vt c
khi lng m = 200g, tn s f = 2Hz. Ly 210 , thi im t1 vt c li x1 =
4cm, th nng ca con lc thi im t2 sau thi im t1 1,25s l :A. 256mJ B.
2,56mJ C. 25,6mJ D. 0,256mJDNG 7. BI TON V LCI. Phng php. Bi ton:Tm
lc tc dng ln nht, nh nht vo im treo hay nn ln sn... Hng dn: + Bc 1:
Xem lc cn tm l lc g?V d hnh bn : dhFuuur + Bc 2: Xt vt thi im t, vt
c li x, p dng nh lut2 Newton dng v hng, ri rt ra lc cn tm.". . .
.dh dhma P F F P ma m g m x (1) + Bc 3: Thay " 2. x x vo (1) ri bin
lun lc cn tm theo li x. Ta c2. . .dhF m g m x + .* 2( ) . . .dhF
Max m g m A + khi x = +A (m)* Mun tm gi tr nh nht ca Fh ta phi so
snh l ( bin dng ca l xo ti v tr cn bng) vA (bin dao ng)- Nul < A
2( ) . . .dhF Min m g m l khix l .- Nu l > A 2( ) . . .dhF Min m
g m A khi x = -A.II. Bi Tp.B i 1 .Treo mt vt nng c khi lng m = 100g
vo u mt l xo c cng k = 20 (N/m). u trn ca l xo c gi c nh. Ly g =
10(m/s2).Nguyn nh Khang11O(VTCB)x(+)PurdhFuuurAPhng Php Gii Ton Vt
L 12a) Tm dn ca l xo khi vt VTCB.b) Nng vt n v tr l xo khng b nin
dng ri th nh cho vt dao ng. B qua mi ma st. Chng t vt m dao ng iu
ho. Vit phng trnh dao ng ca vt. Chon gc thi gian l lc th.c) Tm gi
tr ln nht v nh nht ca lc phc hi v lc n hi ca l xo.B i 2. Mt l xo c
treo thng ng, u trn ca l xo c gi c nh, u di ca l xo treo mt vt m =
100g. L xo c cng k = 25(N/m). Ko vt ra khi VTCB theo phng thng ng v
hng xung di mt on 2cm ri truyn cho n mt vn tc 010. . 3 v (cm/s) hng
ln. Chn gc thi gian l lc truyn vn tc cho vt, gc to l VTCB, chiu dng
hng xung. Ly g = 10(m/s2).210 .a) Vit phng trnh dao ng.b) Xc nh thi
im m vt qua v tr l xo dn 2cm ln u tin.c) Tm ln lc phc hi nh cu b.B
i 3. Cho mt con lc l xo c b tr nh hnh v. L xo c cng k=200(N/m); vt
c khi lng m = 500g.1) T v tr cn bng n vt m xung mt on x0 = 2,5cm
theo phng thng ng ri th nh cho vt dao ng.a) Lp phng trnh dao ng.b)
Tnh lc tc dng ln nht v nh nht m l xo nn ln mt gi .2)t ln m mt gia
trng m0 = 100g. T VTCB n h xung mt on x0 ri th nh. a) Tnh p lc ca
m0 ln m khi l xo khng bin dng. b) m0 nm yn trn m th bin dao ng phi
tho mn iu kin g? Suy ra gi tr ca x0. Ly g =10(m/s2).B i 4.Mt l xo c
cng k = 40(N/m) c t thng ng , pha trn c vt khi lng m = 400g. L xo
lun githng ng.a) Tnh bin dng ca l xo khi vt cn bng. Ly g = 10
(m/s2).b) T VTCB n xung di mt on x0 = 2cm ri bung nh. Chng t vt m
dao ng iu ho. Tnh chu k dao ng.c) Tnh lc tc dng ln nht v nh nht m l
xo nn ln sn.B i 5. Mt l xo k = 100(N/m) pha trn c gn vt khi lng m =
100g. Mt vt khi lng m0 = 400g ri t do t cao h = 50cm xung a. Sau va
chm chng dnh vo nhau v dao ng iu ho. Hy tnh :a) Nng lng dao ng.b)
Chu k dao ng.c) Bin dao ng.d) Lc nn ln nht ca l xo ln sn. Ly g = 10
(m/s2).DNG 8. XC NH THI IM CA VT TRONG QU TRNH DAO NGI. Phng
php.Nguyn nh Khang12hm0mkm0mPhng Php Gii Ton Vt L 12B i ton 1: Xc
nh thi im vt i qua v tr cho trc trn qu o.Hng dn: Gi s phng trnh dao
ng ca vt c dng:.sin( . ) x A t +, trong A, , bit. Thi im vt i qua v
tr c li x0 c xc nh nh sau: 00.sin( . ) sin( . )xx A t x tA + + . t
0sinxA ( . ) sin sin t + Vi ;2 2 1 1 ]. *) Nu vt i qua v tr c li x0
theo chiu dng th :. . ( . ) v A cos t + > 0 . Vy thi im vt i qua
v tr c li x0 c xc nh :
.2. .2 .kt k t k T + + + +(Vi iu kin t > 0; k l s nguyn, T l
chu k dao ng).*) Nu vt i qua v tr c li x0 theo chiu m th : . . ( .
) v A cos t + < 0 . Vy thi im vt i qua v tr c li x0 c xc nh
:
.2. .2 .kt k t k T + + + +(Vi iu kin t > 0; k l s nguyn, T l
chu k dao ng).Ch : Tu theo iu kin c th ca u bi m ly k sao cho ph
hp.B i ton 2: Xc nh khong thi gian ngn nht vt i t v tr c li x1 n v
tr c li x2.Hng dn: + Cch 1: Khi chn thi im ban u t = 0 khng phi l
thi im vt v tr c li x1 th khong thi gian t cn tnh c xc nh t h thc t
= t2- t1 , trong t1, t2 c xc nh t h thc : 11 1 1.sin( . ) sin( .
)xx A t tA + + 1... t 22 2 2.sin( . ) sin( . )xx A t tA + + 2... t
+ Cch 2:Khi chn thi im ban ut = 0 l thi im vt v tr c li x1 v chuyn
ng theo chiu t x1n x2th khong thi gian cn xc nh c xc nh t
phngtrnhsau : 22.sin( . ) sin( . )xx A t x tA + + ... t
+ Cch 3: Da vo mi lin h gia chuyn ng trn u v dao ng iu ho. Khong
thi gian c xc nh theo biu thc : t B i ton 3: Xc nh thi im vt c vn
tc xc nh.Hng dn: Gi svt daongvi phngtrnh.sin( . ) x A t +, vntccavt
cdng: . . ( . ) v A cos t +.Thi im vn tc ca vt l v1c xc nh theo
phng trnh: 11. . ( . ) ( . ).vv A cos t v cos tA + + .*) Nu vt
chuyn ng theo chiu dng : v1 > 0. Nguyn nh Khang13Ax(cm)Ox1x2
Phng Php Gii Ton Vt L 12t 1.vcosA ( . ) cos t cos + . 12. .2. .2t
kt k + ++ +
12..t k Tt k T + +
Ch : - Vi k l s nguyn, t > 0, T l chu k - H thc xc nh t1 ng x
> 0, h thc xc nh t2 ng vi x < 0.*) Nu vt chuyn ng ngc chiu
dng : v1 < 0.t 1.vcosA ( . ) cos t cos + .12. .2. .2t kt k + ++
+ +
12..t k Tt k T + + +
Ch : - Vi k l s nguyn, t > 0, T l chu k - H thc xc nh t1 ng x
> 0, h thc xc nh t2 ng vi x < 0.- xc nh ln th bao nhiu vn tc
ca vt c ln v1 khi chuyn ng theo chiu dng hay chiu m, cn cn c vo v
tr v chiu chuyn ng ca vt thi im ban u t = 0.II. Bi Tp.B i 1. Mt vt
dao ng vi phng trnh :10.sin(2. . )2x t +(cm). Tm thi im vt i qua v
tr c li x = 5(cm) ln th hai theo chiu dng. Li Giicc thi im vt i qua
v tr c li x = 5cm c xc nh bi phng trnh: 110.sin(2. . ) 5 sin(2 )2 2
2x t t + + 2. . .22 65.2. . .22 6t kt k + ++ +(; k Z t > 0)Ta c
: '2. .10. (2 )2v x cos t +. V vt i theo chiu dng nn v > 0 '2.
.10. (2 )2v x cos t +> 0. tho mn iu kin ny ta chn 2. . .22 6t k
+ + 16t k + vi k = 1, 2, 3, 4,... (v t > 0)Vt i qua v tr x = 5cm
ln hai theo chiu dng k = 2. Vy ta c t = 1 1126 6 + (s).B i 2 .Mt vt
dao ng iu ho vi phng trnh :10.sin( . )2x t (cm) . Xc nh thi im vt i
qua v tr c li x = -5 2(cm) ln th ba theo chiu m. Li GiiNguyn nh
Khang14Phng Php Gii Ton Vt L 12Thi im vt i qua v tr c li x = -5
2(cm) theo chiu m c xc nh theo phng trnh sau : 210.sin( . ) 5 2
sin( ) sin( )2 2 2 4x t t . Suy ra .22 4.22 4t kt k + + + ( k Z ) .
Ta c vn tc ca vt l : '.10. ( )2v x cos t V vt i qua v tr c li x =
-5 2(cm) theo chiu m nn v < 0. Vy ta c:'.10. ( )2v x cos t <
0. tho mn iu kin ny ta chn .22 4t k + + 72.4t k + (0,1, 2, 3,... k
; t > 0 ) Vt i qua v tr c li x = -5 2(cm) theo chiu m, ln 3 l :
7 232.24 4t + (s).B i 3 . Mt vt dao ng iu ho vi phng trnh
:10.sin(10. . )2x t +(cm). Xc nh thi im vt i qua v tr c li x = 5cm
ln th 2008.Li GiiThi im vt i qua v tr c li x = 5cm c xc nh t phng
trnh:
110.sin(10. . ) 5 sin(10. . )2 2 2x t t + + 10. . .22 6510. .
.22 6t kt k + ++ + v t > 0 nn ta c130 5kt + vi k = 1, 2, 3,
4,... (1) Hoc 130 5kt +vi k = 0, 1, 2, 3, 4,... (2)+ (1) ng vi cc
thi im vt i qua v tr x = 5cm theo chiu dng ( v > 0 ).
'100 . (10 )2v x cos t +> 0v t > 0+ (2) ng vi cc thi im vt
i qua v tr x = 5cm theo chiu m ( v < 0 ).
'100 . (10 )2v x cos t +< 0v t > 0+ Khi t = 0 10.sin 102x
cm , vt bt u dao ng t v tr bin dng. Vt i qua v tr x = 5cm ln th nht
theo chiu m, qua v tr ny ln 2 theo chiu dng. Ta c ngay vt qua v tr
x = 5cm ln th 2008 theo chiu dng, trong s 2008 ln vt qua v tr x =
5cm th c 1004 ln vt qua v tr theo chiu dng.Vy thi im vt qua v tr x
= 5cm ln th 2008 l :130 5kt +vi k = 1004.
1 1004 6024 1 602330 5 30 30t + (s).B i 4.Mt vt dao ng iu ho c
bin bng 4 (cm) v chu k bng 0,1 (s). a) Vit phng trnh dao ng ca vt
khi chn t = 0 l lc vt i qua v tr cn bng theo chiu dng.b) Tnh khong
thi gian ngn nht vt i t v tr c li x1 = 2 (cm) n v tr x2 = 4 (cm).Li
Gii a) Phng trnh dao ng : Phng trnh c dng :.sin( . ) x A t +Nguyn
nh Khang15Phng Php Gii Ton Vt L 12Trong : A = 4cm, 2 220 ( /
)0,1rad sT .Chn t = 0 l lc vt qua VTCB theo chiu dng, ta c : x0 =
A.sin = 0, v0 = A..cos > 0 0( ) rad . Vy 4.sin(20 . ) x t (cm)
b) Khong thi gian ngn nht vt i t v tr c li x1 = 2 (cm) n v tr x2 =
4 (cm).+ Cch 1:- 114sin(20 . ) 2 sin(20 . )2x x t t 11( )120t s ( v
v > 0 )- 24sin(20 . ) 4 sin(20 . ) 1 x x t t 21( )40t s ( v v
> 0 )Kt lun : Khong thi gian ngn nht vt i t v tr c li x1 = 2
(cm) n v tr x2 = 4 (cm) l : t = t2 t1 = 1 1 1( )40 120 60s .+Cch2:
Chnt =0llcvt i quav tr cli x0=x1=2cmtheochiudng, tac: 0 114.sin( )
2 sin2 6x x x (rad) ( v v > 0 )4.sin(20 . )6x t + (cm).Thi gian
vt i t v tr x0n v tr x = 4cmc xc nh bi phng trnh: 14.sin(20 . ) 4
sin(20. . ) 1 ( )6 6 60x t t t s + + ( v v > 0 )+ Cch 3 : Da vo
mi lin h gia chuyn ng trn u v dao ng iu ho: Da vo hnh v ta c : cos
= 2 14 2 3 (rad).Vy t = 1( )3.20 60s . Bi 5. Mt vt dao ng iu ho
theo phng trnh : 10.sin(10 . ) x t (cm). Xc nh thi im vn tc ca vt c
ln bng na vn tc cc i ln th nht, ln th hai. Li Gii+Tphng
trnh10.sin(10 . ) x t (cm)'100. . (10. . )( / ) v x cos t cm s .
Suy ra vn tc cc i l: . 10 .10 100 ( / )maxv A cm s .+ Khi t = 0, v
> 0 vt bt u chuyn ngt VTCB, theo chiu dng. Ln th nht vt chuyn ng
theo chiu dng v c ln vn tc bng na vn tc cc i. Ln th hai vt chuyn ng
ngc chiu dng.+ Khi vt chuyn ng theo chiu dng, ta c : 1100. . (10. .
) .1002v cos t 1(10. . )2cos t 10. . .2310. . .23t kt k + +( vi ; k
Z t > 0 )130 5kt + vi k = 0, 1, 2, 3, ....(1)
130 5kt + vi k =1, 2, 3, ...... (2)Nguyn nh Khang16O24x(cm)Phng
Php Gii Ton Vt L 12H thc (1) ng vi li ca vt 10.sin(10 . ) x t >
0.H thc (2) ng vi li ca vt 10.sin(10 . ) x t < 0.Do vt bt u
chuyn ng t VTCB theo chiu dng nn ln u tin vn tc ca vt bng na vn tc
cc i thi im, 1( )30t s ( k = 0 ).+ Khi vt chuyn ng ngc chiu dng:
1100. . (10. . ) .1002v cos t 1(10. . )2cos t 210. . .23210. . .23t
kt k + +( vi ; k Z t > 0 )115 5kt +(vi k = 0, 1, 2, 3, ....; t
> 0 )(3)
115 5kt +(vi k =1, 2, 3, ......; t > 0)(4)H thc (3) ng vi li
ca vt 10.sin(10 . ) x t > 0.H thc (4) ng vi li ca vt 10.sin(10 .
) x t < 0.Do vt bt u chuyn ng t VTCB theo chiu dng nn ln th hai
vn tc ca vt c ln bng na vn tc cc i thi im, 1( )15t s ( k = 0 ).B i
6. Mt vt dao ng iu ho theo phng trnh :10.sin(5 . )2x t (cm). Xc nh
thi im vn tc ca vt c ln bng 25 2.(cm/s) ln th nht, ln th hai v ln
th ba.Li Gii- Khi t = 010 x cm . Vt btt u chuyn ng t v tr bin m (
x= -A). Do khi vt chuyn ng theo chiu dng th c ln 1 v ln th 2 vn tc
uc ln 25 2.(cm/s), nhng ln 1 ng vi x < 0, cn ln 2 ng vi x >
0. Ln th 3 vn tc ca vt bng25 2.(cm/s) khi vt chuyn ng theo chiu m.-
Vt chuyn ng theo chiu dng, thi im ca vt c xc nh nh sau:
250. . (5 ) 25 2. (5 )2 2 2v cos t cos t 5 .22 45 .22 4t kt k +
+( k Z ) 30, 4.20t k + (vi k= 0, 1, 2, 3, 4, .....); ng vi x > 0
(1) 10, 4.20t k + (vi k= 0, 1, 2, 3, 4, .....); ng vi x < 0
(2)Vt bt u chuyn ng t v tr bin m nn ln th 1 v ln th 2 vn tc ca vt
bng 25 2.(cm/s) cc thi im tng ng l : 11( ) 0, 05( )20t s s ( theo h
thc (2), ng k = 0 ). 23( ) 0,15( )20t s s ( theo h thc (1), ng k =
0 ).- Vt chuyn ng theo chiu m, thi im ca vt c xc nh nh sau :Nguyn
nh Khang17Phng Php Gii Ton Vt L 12250. . (5 ) 25 2. (5 )2 2 2v cos
t cos t 35 .22 435 .22 4t kt k + +( k Z ) 10, 4.4t k + (vi k= 0, 1,
2, 3, 4, ...; t > 0 ); ng vi x > 0(3) 10, 4.20t k + (vi k= 1,
2, 3, 4, .....; t > 0 ); ng vi x < 0(4)Vy vt bt u chuyn ng t
v tr bin m nn ln th 3 vn tc ca vt bng 25 2.(cm/s) thi im tng ng l :
31( ) 0, 25( )4t s s ( theo h thc (3), ng k = 0 ).DNG 9.XC NH VN
TC, GIA TC TI MT IM TRN QU OI. Phng php 1. xc nh vn tc ti mt im trn
qu o, ta lm nh sau :- Ti v tr vt c li l x, vn tc l v, ta c :.sin(
). . ( )x A tv A cos t + + .sin( ). ( )x A tvAcos t + +Bnh phng hai
v, cng v vi v, ta c: 22 2 2 22vA x v A x + t . - Ch : + v > 0 :
vn tc cng chiu dng trc to . + v < 0 : vn tc ngc chiu dng trc to
.2. xc nh gia tc ti mt im trn qu o, ta p dng cng thc:
2. a x - Ch : + a > 0 : gia tc cng chiu dng trc to . + a <
0 : gia tc ngc chiu dng trc to .II. Bi TpB i 1. Mt vt dao ng iu ho
vi chu k( )10T s v i c qung ng 40cm trong mt chu k. Xc nh vn tc v
gia tc ca vt khi i qua v tr c li x = 8cm theo chiu hng v VTCB. Li
Gii-ADCT: 40104 4sA cm ; 2 220( / )10rad sT -Tac: .sin( ). . ( )x A
tv A cos t + + .sin( ). ( )x A tvAcos t + + Bnh phng hai v, cng v
vi v, ta c: 22 2 2 22vA x v A x + t .- Theo u bi ta c: 2 2 2 220.
10 8 120( / ) v A x cm s ( v v < 0 )Nguyn nh Khang18Phng Php Gii
Ton Vt L 12- Ta c : 2 2 2 2. 20 .8 3200( / ) 32( / ) a x cm s m s .
Du chng t gia tc ngc chiu vi chiu dng trc to , tc l n hng v VTCB.B
i 2.Mt vt dao ng iu ho trn on thng di 10cm v thc hin 50 dao ng
trong 78,5s. Tm vn tc v gia tc ca vt khi n i qua v tr c to x = -3cm
theo chiu hng v VTCB. Li Gii- Bin : A = 1052 2lcm ; Chu k: T = 78,
51, 5750tsn ; Tn s gc: 24( / ) rad sT . Vn tc: 2 2 2 24 5 3 16 /
0,16( / ) v A x cm s m s - Gia tc:2 2 2 2. 4 .( 3) 48( / ) 0, 48( /
) a x cm s m s DNG 10. XC NH QUNG NG I C SAU KHONG THI GIAN CHOI.
Phng php+ Khi pha ban u bng : 0, 2t : - Nu trong khong thi gian t,
s chu k dao ng m vt thc hin c l: n, 12n + , 14n + , 34n + , ( n l s
nguyn ) th qung ng m vt i c tng ng l n.4A, (12n + ).4A, (14n +
).4A, (34n + ).4A, ( A l bin dao ng). - Nu trong khong thi gian t,
s chu k dao ng n m vt thc hin khc vi cc s ni trn th qung ng m vt i
c tnh theo cng thc : s = s1 + s2.Trong s1 l qung ng i dc trong n1
chu k dao ng v c tnh theo mt s trung hp trn, vi n1 nh hn hoc gn n
nht. Cn s2 l qung ng m vt i c trong phn chu k cn li n2, vi n2 = n
n1. tnh s2 cn xc nh li ti thi im cui cng ca khong thi gian cho v ch
n v tr, chiu chuyn ng ca vt sau khi thc hin n1 chu k dao ng. C th:
Nu sau khi thc hin n1 chu k dao ng, vt VTCB v cui khong thi gian t,
vt c li l x th : s2 = x. Nu sau khi thc hin n1 chu k dao ng, vt v
tr bin v cui khong thi gian t, c li x th : s2 = A - x.+ Khi pha ban
u khc 0, 2t :- Nu trong khong thi gian t, s chu k dao ng m vt thc
hin c l:n hoc 12n + , ( n nguyn) th qung ng i c tng ng l: n.4A,
(12n + ).4A- Nu trong khong thi gian t, s chu k dao ng n m vt thc
hin khc vi cc s ni trn th qung ng m vt i c tnh theo cng thc : s =
s1 + s2. Trong s1 l qung ng i dc trong n1 chu k dao ng v c tnh theo
mt s trung hp trn, vi n1 nh hn hoc gn n nht. Cn s2 l qung ng m vt i
c trong phn chu k cn li n2, vi n2 = n n1. tnh s2 cn xc nh li x v
chiu chuyn ng ca vt thi im cui ca khong thi gian cho v ch khi vt i
t v tr x1( sau khi thc hin n1 dao ng ) n v tr c li x th chiu chuyn
ng c thay i hay khng?Nguyn nh Khang19Phng Php Gii Ton Vt L 12Ch :
Tm n ta da vo biu thc sau : tnT .II. Bi Tp.B i 1. Mt cht im dao ng
iu ho vi phng trnh: 5.sin(2 . ) x t (cm).Xc nh qung ng vt i c sau
khong thi gian t(s) k t khi vt bt u dao ng trong cc trng hp sau :
a) t = t1 = 5(s).b) t = t2 = 7,5(s). c) t = t3 = 11,25(s). Li Gii-
T phng trnh : 5.sin(2 . ) x t 22 ( / ) 1( )2rad s T s .a) Trong
khong thi gian t1= 5s, s dao ng m vt thc hin c l :1551tnT (chu k).
Vy qung ng m vt i c sau khong thi gian t1 = 5l : s = n.4A = 5.4.5 =
100cm = 1m.b) Trong khong thi gian t2 = 7,5s, s dao ng m vt thc hin
c l :27, 57, 51tnT (chu k). Vy qung ng m vt i c sau khong thi gian
t2 =7, 5s l : s =7,5.4A =7,5 . 4 . 5 = 150cm = 1,5 m.c) Trong khong
thi gian t3 = 11,25s, s dao ng m vt thc hin c l :311, 2511, 251tnT
(chu k). Vy qung ng m vt i c sau khong thi gian t3 =11, 25s l : s
=11,25.4A =11,25 . 4 . 5 = 225cm = 2,25 m.B i 2 .Mt cht im dao ng
iu ho vi phng trnh:10.sin(5 . )2x t + (cm).Xc nh qung ng vt i c sau
khong thi gian t(s) k t khi vt bt u dao ng trong cc trng hp sau :
a) t = t1 = 1(s).b) t = t2 = 2(s). c) t = t3 = 2,5(s). Li GiiT phng
trnh :10.sin(5 . )2x t +5 ( / ) rad s 20, 45T s a) Trong khong thi
gian t1 = 1s, s dao ng m vt thc hin c l :112, 50, 4tnT (chu k). Vy
qung ng m vt i c sau khong thi gian t1 = 1(s) l : s = n.4A = 2,5 .
4 .10 = 100cm = 1m.b) Trong khong thi gian t2 = 2s, s dao ng m vt
thc hin c l :2250, 4tnT (chu k). Vy qung ng m vt i c sau khong thi
gian t2 =2s l : s =5.4A =5 . 4 . 10 = 200cm = 2 m.c) Trong khong
thi gian t3 = 2,5, s dao ng m vt thc hin c l :32, 56, 250, 4tnT
(chu k). Vy qung ng m vt i c sau khong thi gian t3 =2,5s l : s
=11,25.4A =6,25 . 4 . 5 = 250cm = 2,5 m.B i3 .Mtchtimdaongiuho vi
phngtrnh: 10.sin(5 . )6x t + (cm). Xc nh qung ng vt i c sau khong
thi gian t(s) k t khi vt bt u dao ng trong cc trng hp sau : a) t =
t1 = 2(s).b) t = t2 = 2,2(s). c) t = t3 = 2,5(s). Li GiiT phng trnh
:10.sin(5 . )6x t +5 ( / ) rad s 20, 45T s a) Trong khong thi gian
t1 = 2s, s dao ng m vt thc hin c l :Nguyn nh Khang20Phng Php Gii
Ton Vt L 121250, 4tnT (chu k). Vy qung ng m vt i c sau khong thi
gian t1 = 2(s) l : s = n.4A = 5 . 4 .10 = 200cm = 2m.b) Trong khong
thi gian t2 = 2,2s, s dao ng m vt thc hin c l :22, 25, 50, 4tnT
(chu k). Vy qung ng m vt i c sau khong thi gian t2 =2s l : s =5,5 .
4A =5,5 . 4 . 10 = 220cm = 2,2 m.c) Trong khong thi gian t3 = 2,5,
s dao ng m vt thc hin c l :32, 56, 250, 4tnT (chu k). - thi im t3 =
2,5(s), li ca vt l: 210.sin(5 .2, 5 ) 10.sin 5 3( )6 3x cm + Nh vy
sau 6 chu k dao ng vt tr v v tr c li 02Ax theo chiu dng v trong
0,25 chu k tip theo, vt i t v tr ny n v tr bin x = A, ri sau i chiu
chuyn ng v i n v tr c li 5 3( ) x cm . Qung ng m vt i c sau 6,25
chu k l: s = s1 + s2 = 6 . 4. 10 + ( A x0) + ( A x) = 246,34(cm).B
i 4 Mt vt daongiuhodctheotrcOx, xungqu8anhVTCBx=0. Tnsdaong 4( / )
rad s . Ti mt thi im no , li ca vt l x0 = 25cm v vn tc ca vt l v0=
100cm/s. Tm li x v vn tc ca vt sau thi gian32, 4( )4t s . S : x =
-25cm, v = -100cm/s.B i 5. Mt vt dao ng iu ho theo phng trnh :
.sin( . ) x A t +. Xc nh tn s gc, bin A ca dao ng. Cho bit, trong
khong thi gian 1/60 (s) u tin, vt i t v tr x0 = 0 n v tr x = 32A
theo chiu dng v ti im cch VTCB 2(cm) vt c vn tc40 3 (cm/s). S : 20
( )rads ,A= 4(cm).B i 6.Mt vt dao ng iu ho i qua VTCB theo chiu dng
thi im ban u. Khi vt c li l 3(cm) th vn tc ca vt l8 (cm/s), khi vt
c li l 4(cm) th vt c vn tc l6 (cm/s). Vit phng trnh dao ng ca vt ni
trn. S : 5.sin(2 . ) x t cm . DNG 11. H MT L XO C LIN KTRNG RCI.
Phng php- p dng nh lut bo ton v cng: Cc my c hc khng cho ta c li v
cng, tc l c li bao nhiu ln v lc th thit by nhiu ln v ng i- V d :
Rng rc, n by, mt phng nghing,...II.Bi tpB i 1.Cho hai c h b tr nh
hnh v. L xo c cng k = 20(N/m), vt nng c khi lng m = 100g. B qua lc
ma st, khi lng ca rng rc, khi lng dy treo Nguyn nh
Khang21Pur1TurdhFuuur2Tuur3TuurO(VTCB)Pur1TurIdhFuuur2Tuura)b)Phng
Php Gii Ton Vt L 12( dy khng dn ) v cc l xo l khng ng k. 1. Tnh dn
ca mi l xo khi vt VTCB. Ly g = 10(m/s2). 2. Nng vt ln v tr sao cho
l xo khng bin dng, ri th nh cho vtdao ng. Chng minh vt m dao ng iu
ho. Tm bin , chu k ca vt. Li Giia) Hnh a:Chn HQC l trc to Ox, O
trng vi VTCB ca m, chiu dng hng xung.- Khi h VTCB, ta c:+ Vt m: 10
P T + ur ur.+ im I: 20dhT F + uur uuur. Chiu ln HQC, ta c10 P T
(1).20dhF T (2). V l xo khng dn nn T1 = T2. T (1) v (2), ta c : P =
Fh (*). 0,1.10. . 0, 05 520m gm g k l l m cmk .- Khi h thi im t, c
li x, ta c:+ Vt m : 1. P T ma + ur ur r+ im I: 2.dh IT F m a + uur
uuur r. V mI = 0 nn ta c: 1. P T ma (3).20dhF T (4). . . ( ) .dhP F
ma m g k x l ma + (**)Thay (*) vo (**) ta c: " ". . . 0kk x m x x
xm + . t 2 " 2. 0kx xm + . C nghim dng .sin( ) x A t + H vt dao ng
iu ho, vi tn s gc km .- Khi nngvt ln v tr saochol xokhngbin dng, ta
suyra A=5cm. Chukdaong 2 0,12 2 . 0, 314 220mTk (s).b) Hnh b:- Khi
h VTCB, ta c:+ Vt m: 10 P T + ur ur.+ Rng rc: 2 30dhT T F + + uur
uur uuur. Chiu ln HQC, ta c : 10 P T (5).3 20dhF T T + + (6). V l
xo khng dn nnT0 = T3 = T1 = T2. T (6) ta suy ra 02.dhF T 02dhFT .
Thayvophngtrnhs(5)tac:2. .0 2. . . 0,1 102 2dh dhF F m gP P m g k l
l m cmk . (***)-Khi h thi im t, c li x, ta c:+ Vt m : 1. P T ma +
ur ur r+ Rng rc: 2 3.dh rrT T F m a + + uur uur uuur r. Chiu ln
HQC, ta c : 1. P T ma (7)V mrr= 0 nn ta c:3 20dhF T T + + (8). V l
xo khng dn nn T0= T3= T1= T2. T (8) ta suy ra 02.dhF T thay vo (7)
ta c: "1. . . .( ) .2 2 2dhF xP ma m g k l m x + ( V theo nh lut bo
ton Nguyn nh Khang22Phng Php Gii Ton Vt L 12cng ta c, khi vt m i
xung mt on l x th l xo dn thm mt on x/2 ). Thay (***) vo ta c: "
".. . 04 4.k x km x x xm + . t 24km " 2. 0 x x + . Vy vt m dao ng
iu ho. Bin dao ng A=20cm; chu k dao ng T = 2 2 4 4.0,12 . 2 0, 628
2204mk km (s).B i 2.Qu cu khi lng m1= 600g gn vo l xo c cng k =
200(N/m). Vt nng m2= 1kg ni vi m1bng si dy mnh , khng dn vt qua rng
rc. B qua mi ma st ca m1v sn, khi lng rng rc v l xo l khng ng k.a)
Tm dn ca l xo khi vt cn bng. Ly g = 10(m/s2).b) Ko m2 xung theo
phng thng ng mt on x0 = 2cm ri bung nh khng vn tc u. Chng minh m2
dao ng iu ho. Vit phng trnh dao ng.B i 3.Cho mt h vt dao ng nh hv.
L xo v rng rc khi lng khng ng k. cng ca l xo k = 200 N/m, M = 4 kg,
m0=1kg. Vt M c th trt khng ma st trn mt phng nghing gc nghing =
300.a) Xc nh bin dng ca l xo khi h cn bng.b) T VTCB, ko M dc theo
mt phng nghing xung dimt onx0=2,5cmri thnh. CMhdaongiuho. Vit phng
trnh dao ng. Ly g = 10 m/s2, 2 = 10.B i 4 : Mt l xo c cng k = 80
N/m, l0=20cm, mt u c nh u kia mc vo mt vt C khi lng m1 = 600g c th
trt trn mtmt phng nm ngang. Vt C c ni vi vt D c khi lng m2= 200g
bng mt si dy khng dn qua mt rng rc si dy v rng rc c khi lng khng ng
k. Gi vt D sao cho l xo c di l1= 21cm ri th ra nh nhng. B qua mi ma
st, ly g = 10 m/s2, 2 = 10.a) Chng minh h dao ng iu ho v vit phng
trnh dao ng.b) t h thng l xo, vt C cho trn mt phng nghing gc = 300.
Chng minh h dao ng iu ho v vit phng trnh dao ng.DNG 12. IU KIN HAI
VT DAO NG CNGGIA TC I. Phng php- Trng hp 1.Khi m0 th ln m v kch
thch cho h dao ng theo phng song song vi b mt tip xc gia hai vt. m0
khng b trt trn m th lc ngh ma st cc i m m tc dng m0 trong qu trnh
dao ng phi nh hn hoc bng lc ma st trt gia hai vt.fmsn (Max) <
fmst 20 0 0 0. . . . . . . m a m g m x m g
20 0. . . . m A m g Trong : l h s ma sttrt.- Tr ng hp 2.Khi m0 t
ln m v kch thch cho h dao ng theo phng thng ng. m0 khng ri khi m
trong qu trnh dao ng th: amax 2. g A g II. Bi TpNguyn nh
Khang23m0Mkm1m2m1m2 kdhFuuurTurTurTurTurPurmAPhng Php Gii Ton Vt L
12B i 1. Cho c h dao ng nh hnh v, khi lng ca cc vt tng ng l m =
1kg, m0 = 250g, l xo c khi lng khng ng k, cng k = 50(N/m). Ma st
gia m v mt phng nm ngang khng ng k. H s ma st gia m v m0l 0, 2 . Tm
bin dao ng ln nht ca vt m m0 khng trt trn b mt ngang ca vt m. Cho g
= 10(m/s2), 210 . Li Gii- Khi m0 khng trt trn b mt ca m th h hai vt
dao ng nh l mt vt( m+m0 ). Lc truyn gia tc cho m0 l lc ma st ngh
xut hin gia hai vt.
20 0. . .msnf m a m x . Gi tr ln nht ca lc ma st ngh l : 20( ) .
.msnf Max m A (1)- Num0trt trnbmt camth lcmast trt xut hingiahai vt
llcmast trt : 0. .mstf m g (2)- m0 khng b trt trn m th phi c: 20 0(
) . . . .msn mstf Max f m A m g 2.gA ; m 20km m + nn ta c : 0. . 0,
05 5 .m mA g A m A cmk+ Vy bin ln nht ca m m0 khng trt trn m l Amax
= 5cm.B i 2.Mt vt c khi lng m = 400g c gn trn mt l xo thng ng c cng
k = 50(N/m). t vt m c khi lng 50g ln trn m nh hnh v. Kch thch cho m
dao ng theo phng thng ng vi bin nh. B qua sc cn ca khng kh. Tm bin
dao ng ln nht ca m m khng ri khi m trong qu trnh dao ng. Ly g = 10
(m/s2).
L i Gii m khng ri khi m trong qu trnh dao ng th h ( m+m) dao ng
vi cng gia tc. Ta phi c:amax 2. g A g 2( ').0, 09g m m gA A A mk +
9 9maxA cm A cm . DNG 13.BI TON V VA CHMI. Phng php- nh lut boton
ng lng :p const ur 1 2 3...np p p p Const + + + + uur uur uur uur.
(iu kin p dng l h kn)- nh lut bo ton c nng : E = const E + Et =
const. (iu kin p dng l h kn, khng ma st)- nh l bin thin ng nng : d
ngoailucE A 2 22 1 2 11 1. . . .2 2d d ngoailuc ngoailucE E A mv mv
A .- Ch : i vi va cham n hi ta c : 2 2 2 22 2 1 1 2 2 1 11 1 1 1. .
. . . . ' . . '2 2 2 2m v m v m v m v + +Nguyn nh
Khang24mm0kmmkPhng Php Gii Ton Vt L 12II. Bi TpB i 1 . C h dao ng
nh hnh v gm mt vt M = 200g gn vo l xo c cng k, khi lng khng ng k.
Vt M c th trt khng ma st trn mt ngang. H trng thi cn bng ngi ta bn
mt vt m = 50g theo phng ngang vi vn tc v0 = 2(m/s) n va chm vi
M.Sau va chm, vt M dao ng iu ho, chiu di cc i v cc tiu ca l xo l
28cm v 20cm.a) Tnh chu k dao ng ca M.b) Tnh cng k ca l xo. Li Gii
a) Tm chu k dao ng:- p dng LBTL:0. . . mv mv M V +uur r ur; trong ;
v Vr ur l vn tc ca m v M ngay sau va chm. Phng trnh v hng: 0. . .
mv mv M V + 0 0.( ) . .Mm v v M V v v Vm (1)- p dng LBTCN:
2 2 2 2 2 2 2 2 20 0 01 1 1. . . . . . .( ) . ( ) .2 2 2Mmv mv
MV m v v MV v v Vm + (2)Ly (2) chia cho (1) ta c: v0 + v =V (3)Ly
(1) cng (3), ta c: 002. .2. . 0, 8( / )mv M mv V V m sm M m+ +.Mt
khc ta c :min4 .2maxl lA cm Vn tc ca M ngay sau va chm l vn tc cc i
trong dao ng ca vt M, ta c 2 2 . 2 .4. . 0, 314( )80AV A A T sT V
.b) Tm cng k ca l xo:22 224.. . 80( / )kk M M N mM T .B i 2. Mt ci
a khi lng M = 900g t trn l xo c cng k = 25(N/m).Mt vt nh m = 100g
ri khng vn tc ban u t cao h = 20(cm) ( so vi a) xung a v dnh vo a.
Sau va chm h hai vt dao ng iu ho.1. Vit phng trnh dao ng ca h hai
vt, chn gc to l VTCB ca h vt, chiu dng hng thng ng t trn xung, gc
thi gian l lc bt u va chm. Ly g = 10(m/s2).2. Tnh cc thi im m ng
nng ca hai vt bng ba ln th nng ca l xo.Ly gc tnh th nng ca l xo l
VTCB ca hai vt.Li Gii1. Chn mt phng i qua a lm mc tnh th nng, ta
c:Gi v0 l vn tc ca m ngay trc va chm, p dng LBTCN, ta c 200.. . 2.
. 2( / )2mvm g h v g h m s Do va chm l va chm mm nn ngay sau khi va
cham c h chuyn ng vi vn tc v ; p dng LBTL, ta c: 00.. ( ). 20( /
)mvmv M m v v cm sM m + +.Khi h VTCB, h nn thm mt on l: . . 4( )mgm
g k l l cmk Nguyn nh Khang25Mm0vuurkmMkhPhng Php Gii Ton Vt L
12Phng trnh c dng:.sin( ) x A t +; vi 5( / )krad sM m + thi im ban
u, t = 0 00.sin 4. . 20 /x A cmv A cos cm s ; 4 24rad A cm . 4
2.sin(5 )4x t cm Nu vit phng trnh theo hm cosin ta c: ( ) x Acos t
+ thi im ban u, t = 0 00. 4. .sin 20 /x Acos cmv A cm s 3; 4 24rad
A cm .
34 2. (5 )4x cos t cm +2. Tm cc thi im m E= 3Et: Ta c E = E+ Et=
21. .2k A m E= 3.Etnn thay v ta c: 4Et= E 2 21 14. . . . .2 2 2Ak x
k A x t 3 4 24 2. (5 )4 2x cos t + t 3 1(5 )4 2cos t+ tKhi 3 1(5 )4
2cos t+ 35 .24 335 .24 3t nt n + ++ +5 2.60 513 2.60 5t nt n + + vi
1, 2, 3, 4,...1, 2, 3, 4, 5,...nn Khi 3 1(5 )4 2cos t+ 3 25 .24 33
25 .24 3t nt n + ++ +2.60 517 2.60 5t nt n + + vi 1, 2, 3, 4,
5,...1, 2, 3, 4, 5,...nn B i 3.Mt ci a nm ngang, c khi lng M =
200g, c gn vao u trn ca mt l xo thng ng c cng k = 20(N/m). u di ca
l xo c gi c nh. a c th chuyn ng theo phng thng ng. B qua mi ma st v
sc cn ca khng kh.1. Ban u a VTCB. n a xung mt on A = 4cm ri th cho
a dao ng t do. Hy vit phng trnh dao ng ( Ly trc to hng ln trn, gc
to l VTCB ca a, gc thi gian l lc th).2. a ang nm VTCB, ngi ta th mt
vt c khi lng m = 100g, t cao h = 7,5cm so vi mt a. Va chm gia vt v
a l hon ton n hi. Sau va chm u tin vt ny ln v c gi khng cho ri xung
ana. Ly g = 10(m/s2) a) Tnh tn s gc dao ng ca a.b) Tnh bin A dao ng
ca a. c) Vit phng trnh dao ng ca a.Li Gii1. Phng trnh dao ng c dng
: . ( ) x Acos t +. Trong : 2010( / )0, 2krad sM ; Nguyn nh
Khang26Phng Php Gii Ton Vt L 12theo iu kin ban u ta c: t = 0 00. 4.
.sin 0x Acos cmv A 40sin 0cosAp ; 4 A cm . Vy ta c 4. (10 ) 4 (10 )
x cos t cos t cm + .2. Gi v l vn tc ca m trc va chm; v1, V l vn tc
ca m v M sau va chm.Coi h l kn, p dng LBTL ta c: 1. . .t sp p mv mv
M V +uur uur r ur ur. chiu ln ta c:-m.v = m.v1 M.V 1.( ) . m v v M
V + (1)Mt khc ta c: p dng LBTCN : m.g.h = m.222. .2vv g h (2)Do va
chm l tuyt i n hi nn: 2 2 21. .2 2 2mv mv MV +(3)Gii h (1), (2),
(3), ta c : 1, 2( / ) v m s v 0,8( / ) V m s p dng LBTCN trong dao
ng iu ho : E = E + Et ( Et = 0 ) nn E = E2 21 1. . ' . . ' 0.082 8,
22 2k A M V A m cm .3. Phng trnh dao ng ca a c dng : '. ( ) x A cos
t +trong 10( / ) rad s ; A = 8,2cm.Ti thi im ban u t = 0 000 '.'
.sinx A cosv V A 2' 8, 2radA cm . Vy phng trnh ca a l :8, 2. (10
)2x cos t cm + .DNG 14.BITON V DAO NG CA VT SAU KHIRIKHIGI I. Phng
php- Qung ng S m gi i c k t khi bt u chuyn ng n khi vt ri khi gi
bng phn tng bin dng ca l xo trong khong thi gian . Khong thi gian t
lc gi bt u chuyn ng Nguyn nh Khang27Phng Php Gii Ton Vt L 12n khi
vt ri khi gi c xc nh theo cng thc : 21 22SS at ta ( a l gia tc ca
gi ) (1)- Vn tc ca vt khi rikhi gi l :2 . v a S (2)- Gi 0l l bin
dng ca l xo khi vt VTCB ( khng cn gi ),l l bin dng ca l xo khi vt
ri gi . Li x ca vt thi im ri khi gi l 0x l l - Ta c22 22vx A+ II.
Bi Tp.B i 1.Con lc l xo gm mt vt nng c khi lng m = 1kg v mt l xo c
cng k = 100N/m, c treo thng ng nh hnh v. Lc u gi gi D sao cho l xo
khng bin dng. Sau cho D chuyn ng thng ng xung di nhanh dn u vi gia
tc a = 2m/s2.1. Tm thi gian k t khi D bt u chuyn ng cho ti khi m bt
u ri khi D.2. CMRsau khirikhiDvt mdao ng iu ho. Vit phng trnh dao
ng, chiu dng xung di, gc thi gian l lc vt m bt u kri khi D. Ly g =
10m/s2. B qua mi ma st v khi lng ca l xo.Li Gii1. V gi D sao cho l
xo khng bin dng nn khi D chuyn ng xung di th vt m cng chuyn ng xung
di vi cng vn tc v gia tc ca D. Gi s D i c qung ng l S th m ri khi
D. Lc l xo cng dn mt on S. p dng L II Niu Tn ta c : .dhP F ma + ur
uuur r
( )0, 08 8m g amg kS ma S m cmk Mt khc ta c : 21 2. 0, 282SS a t
t sa 2. Chng minh M dao ng iu ho:- xt m VTCB (khng cn gi ) 00dhP F
+ ur uuuur 0 00 0,1 10 .mgmg k l l m cmk (1)- xt vt m thi im t, c
li l x: .dhP F ma + ur uuur r 0( ) mg k l x ma + 0mg k l kx ma ( 2)
Thay (1) vo (2) ta c: 2" 0 " . 0kx x x xm + + vi km . ( ) x Acos t
+ Vy m dao ng iu ho. Ta c10( / )krad sm .Khi ri khi gi vt m c vn tc
l 02 0, 4 2( / ) 40 2( / ) v aS m s cm s thi im ri gi vt m c li x0
so vi gc to . 0 0( ) 2 x l S cm Bin dao ng ca vt l : A2 = 22 00
2vx+ 6 A cm .Nguyn nh Khang28DkmPhng Php Gii Ton Vt L 12Khi t = 0
002 .. .sinx Acosv A 240 2sin10cosAA tan 2 2 .B i 2. Con lc l xo gm
vt c khi lng m = 1kg v l xo c cng k = 50N/m c treo nh hnh v. Khi gi
D ng yn th l xo dn mt on 1cm. Cho D chuyn ng thng ng xung di nhanh
dn u vi gia tc a = 1m/s2, v vn tc ban u bng khng. B qua mi ma st v
sc cn , ly g = 10m/s2.1. xc nh qung ng m gi i c k t khi bt u chuyn
ng n thi im vt ri khi gi .2. Sau khi ri khi gi , vt m dao ng iu ho.
Tnh bin dao ng ca vt. Li gii1. Khi ri khi gi , l xo c bin dng l l .
thi imvt ri khi gi , ta c: .( ). . 0, 09 9dhm g aP F ma mg k l ma l
m cmk+ ur uuur rKhi gi bt u chuyn ng th l xo dn mt on 0l 1cm, do
qung ng i c ca gi k t khi bt u chuyn ng cho ti khi vt ri gi l:09 1
8 S l l cm .2. Sau khi ri khi gi , vt mdao ng iu ho. Ti VTCB l xo
dn mt on l: ' 0,1 10mgl m cmk thi im vt ri khi gi , vt c li l :0( '
) 1 x l l cm Khi ri khi gi , vt c vn tc l: 02 40 / v aS cm s Tn s
gc ca dao ng l: 5 2( / )krad sm Vy bin dao ng l:22 00 233vA x cm +
dng 15tng hp hai dao ng iu ho cng phng, cng tn sNguyn nh
Khang29DkmPhng Php Gii Ton Vt L 12I. Phng php- Cho hai dao ng cng
phng, cng tn s:
1 1 1. ( ) x A cos t + v 2 2 2. ( ) x A cos t +- Dao ng tng hp c
dng :. ( ) x Acos t +Trong A, c xc nh theo cng thc sau: 2 2 21 2 1
2 1 22. . . ( ) A A A A A cos + + ; 1 1 2 21 1 2 2.sin .sintan. .A
AA cos A cos ++- Ch : + C th tm phng trnh dao ng tng hp bng phng
php lng gic + Nu hai dao ng cng pha: A = A1 + A2 + Nu hai dao ng
ngc pha: A = 1 2A A .II. Bi TpB i 1. Hai dao ng c cng phng, cng tn
s f = 50Hz, c bin A1 = 2a, A2 = a. Cc pha ban u 1 2( ); ( )3rad rad
.1. Vit phng trnh ca hai dao ng .2. Tm bin v pha ban u ca dao ng
tng hp. V trn cng mt gin vc t cc vc t 1 2; ; A A Auur uur ur. Li
Gii1. Phng trnh dao ng l: 12 . ( 100 )3x a cos cm + ; 2. (100 ) x a
cos cm +.2. Ta c: 2 2 2 2 2 21 2 1 2 1 222. . . ( ) 4 4 . ( )3A A A
A A cos a a a cos + + + + 2 2 2 25 2 3 3 A a a a A a cm .Pha ban u
ca dao ng tng hp l: 1 1 2 21 1 2 2.sin .sintan. .A AA cos A cos
++
2 .sin .sin33tan ( )0 22 .cos .cos3a aarada a + +.B i 2. Cho hai
dao ng c phng trnh:1 1 2 23sin( ); 5sin( ) x t x t + +Hy xc nh phng
trnh v v gin vc t ca dao ng tng hp trong cc trng hp sau:1. Hai dao
ng cng pha.2. Hai dao ng ngc pha.3. Hai dao ng lch pha mt gc 2 ( xc
nh pha ban u ca dao ng tng hp ph thuc vo 1 2; ).B i 3Cho hai dao ng
cng phng, cng tn s, c cc phng trnh dao ng l : 1 23sin( )( ); 4sin(
)( )4 4x t cm x t cm + . Tm bin ca dao ng tng hp trn?B i 4.Hai dao
ng c iu ho, cng phng, cng tn s gc50 / rad s , c bin ln lt l 6cm v
8cm, dao ng th hai tr pha hn dao ng th nht l 2rad. Xc nh bin ca dao
ng tng hp. T suy ra dao ng tng hp.DNG 16.HIN TNG CNG HNG C HCNguyn
nh Khang30O P2 P1PxMM2M1Phng Php Gii Ton Vt L 12I. Phng php H dao
ng c tn s dao ng ring l f0, nu h chu tc dng ca lc cng bc bin thin
tun hon vi tn s f th bin dao ng ca h ln nht khi: f0 = fII. Bi TpB i
1. Mt chic xe gn my chy trn mt con ng lt gch, c cch khong 9m trn ng
li c mt rnh nh. Chu k dao ng ring ca khung xe my trn l xo gim xc l
1,5s. Hi vi vn tc bng bao nhiu th xe b xc mnh nht.Li GiiXe my b xc
mnh nht khi f0 = f 0T T m T = s/v suy ra v = s/T = 9/1,5 = 6(m/s) =
21,6(km/h).B i 2.Mt ngi xch mt x nc i trn ng, mi bc i c 50cm. Chu k
dao ng ca nc trong x l 1s. Ngi i vi vn tc no th nc trong x b snh
nhiu nht./s : v = 1,8km/hB i 3.Mt hnh khch dng mt si dy cao su
treomt ti xch ln trn toa tu ngay v tr pha trn mt trc bnh xe ca tu
ho. Khi lng ti xch l 16kg, h s cng ca dy cao su 900N/m, chiu dica
mi thanh ray l 12,5m, ch ni hai thanh ray c khe nh. Tu chy vi vn tc
bng bao nhiu th ti xch dao ng mnh nht?/s:v = 15m/s=54km/hB i 4. Mt
con lc n c di l = 30cm c treo trong toa tu ngay v tr pha trn trc ca
bnh xe.Chiu di ca mi thanh ray l 12,5m. Vn tc tu bng bao nhiu th
con lc dao ng mnh nht?/s : v = 41km/h DNG 17.DAO NG CA CON LC L XO
TRONG TRNG LC LI. Phng php* Lc l l lc y Acsimet.AF DV g uur ur- Vt
VTCB : 0 0dh A dh AP F F P F F + + ur uuur uur0. . . 0 mg k l S h
Dg (1)- Xt vt thi im t, c li x: dh A dh AP F F ma P F F ma + + ur
uuur uur r0( ) ( ). . " mg k l x S h x D g mx + + 0. . ( ) " mg k l
S h Dg x k SDg mx + Thay (1) vo ta c:" . 0k SDgx xm++ C nghim dng.
( ) x Acos t +. Vy vt m dao ng iu ho vi tn s gc
k SDgm+*Lc l l lc qun tnh. .qtF ma uur rtrong h quy chiu khng
qun tnh ngoi lc n hi ca l xo, trng lc tc dng vo vt, vt cn chu tc
dng ca lc qun tnh. Du - cho tabit lc qun tnh lun hng ngc vi gia tc
ca chuyn ng.* Lc ma st..mstF N II. Bi TpB i 1. Mt vt nng c dng hnh
tr c khi lng m = 0,4kg, chiu cao h = 10cm, tit din S = 50cm2, c
treo vo mt l xo c cng k = 150N/m. Khi cn bng, mt na vt b nhng chm
trong cht lng c khi lng ring D = 103kg/m3. Ko vt theo phng thng ng
xung di mt on l 4cm ri th nh cho vt dao ng. B qua sc cn. Ly g =
10m/s.1. Xc nh bin dng ca l xo ti VTCB.2. Chng minh vt dao ng iu
ho. Tnh chu k dao ng ca vt.Nguyn nh Khang31PurdhFuuurAFuurk1 m k2
Phng Php Gii Ton Vt L 123. Tnh c nng ca vt.B i 2. Treo con lc l xo
gm mt vt nng c khi lng m = 200g vo l xo c cng k = 80N/m v chiu di t
nhin l0 = 24cm trong thang my. Cho thang my chuyn ng ln trn nhanh
dn u vi gia tc a = 2m/s2. Ly g = 10m/s2.1.Tnh bin dng ca l xo ti
VTCB.2. Kch thch cho vt dao ng vi bin nh theo phng thng ng. Chng
ming m dao ng iu ho. Tnh chu k ca dao ng. C nhn xt g v kt qu?B i
3.Mt con lc l xo gm mt vt nng c khi lng m = 250g gn vo l xo c cng k
= 100N/m v chiu di t nhin l0 = 30cm. Mt u l xo treo vo thang my.
Cho thang my chuyn ng nhanh dn u ln trn vi vn tc ban u bng khngv
gia tc a th thy rng l xo c chiu di l l1 = 33cm.1. Tnh gia tc a ca
thang my. Ly g = 10m/s2.2. Ko vt nng xung di n v tr sao cho l xo c
chiu di l2 = 36cm ri th nh nhng cho dao ng iu ho. Tnh chu k v bin
ca con lc.B i 4Mt vt c khi lng m c gn vo mt l xo c cng kv khi lng l
xo khng ng k. Ko vt ri VTCB dc theo trc ca l xo mt on a ri th nh
nhng cho dao ng. H s ma st gia vt m v mt phng nm ngang l khng i.
Gia tc trng trngl g. B qua lc cnca khng kh. Tnh thi gian thc hin
dao ng u tin ca vt.B i 5. Gn mt vt c khi lng m = 200g vo l xo c cng
k = 80N/m. Mt u l xo c gi c nh. Ko m khi VTCB mt on 10cm dc theo
trc ca l xo ri th nh nhng cho vt dao ng. Bit h s ma st gia m v mt
nm ngang l = 0,1. Ly g = 10m/s2.1. Tm chiu di qung ng m vt i c cho
n khi dng li.2. Chng minh rng gim bin dao ng sau mi mt chu k l mt s
khng i.3. Tm thi gian dao ng ca vt. Li gii1. khi c ma st vt dao ng
tt dn cho n khi dng li. C nng b trit tiu bi cng ca lc ma st. Ta
c:
21. . .2mskA F s mg s 2 2. 80.0,122 . 2.0,1.0, 2.10k As mmg 2.Gi
s ti thi im vt ang v tr c bin A1. Sau na chu k , vt n v tr cbin A2.
S gim bin l do cng ca lc ma st trn on ng (A1 + A2) lm gim c nng ca
vt. Ta c: 2 21 2 1 21 1. ( )2 2kA kA mg A A +1 22 .mgA Ak . Lp lun
tng t, khi vt i t v tr bin A2 n v tr c bin A3, tc l na chu k tip
theo th:2 32 .mgA Ak . gim bin sau mi mt chu k l: 1 2 2 34 .( ) (
)mgA A A A Ak + = Const. pcm3. gim bin sau mi mt chu k l: 0, 01 1 A
m cm S chu k thc hin l10AnA chu k. Vy thi gian dao ng l: t = n.T =
3,14sDNG 18. DAO NG CA MT VT GN VIH HAI L XOI. Phng phpNguyn nh
Khang32k1 m k2 k1 m k2 Phng Php Gii Ton Vt L 12 A. H hai l xo cha c
lin kt. t vn : Hai l xo c chiu di t nhin L01 v L02. Hai u ca l xo
gn vo 2 im c nh A v B. Hai u cn li gn vo 1 vt c khi lng m. Chng
minh m dao ng iu ho, vit phng trng dao ng,...* Trng hp 1. AB = L01
+ L02.( Ti VTCB hai l xo khng bin dng )Xt vt m thi im t c li l x: 1
2.dh dhma F F +r uuur uuuur. Chiu ln trc Ox, ta c:1 2 1 2. . ( ) ma
k x k x x k k + 1 21 2( ) 0 " . 0k kma x k k x xm+ + + + . t2 1 2k
km+ . Vytac: 2" . 0 x x + Cnghiml. ( ) x Acos t +. Vyvt
mdaongiuhovi tnsgcl 1 2k km+* Trng hp 2. AB > L01 + L02 ( Trong
qu trnh dao ng hai l xo lun lun b dn ).- Cch 1: Gi 1l v 2l ln lt l
dn ca hai l xo ti VTCB + Xt vt m VTCB: 0 1 0 20dh dhF F +uuuur
uuuuur. Chiu ln trc Ox, ta c2 2 1 1. . 0 k l k l (1) + Xt vt m thi
im t, c li x: 1 2.dh dhma F F +r uuur uuuurChiu ln trc Ox: 2 1 2 2
1 1" ( ) ( )dh dhma F F mx k l x k l x + (2)Thay (1) vo (2) ta c:1
2 1 2. . ( ) ma k x k x x k k +1 21 2( ) 0 " . 0k kma x k k x xm+ +
+ + . t2 1 2k km+ . Vytac:2" . 0 x x + Cnghiml . ( ) x Acos t +. Vy
vt m dao ng iu ho vitn s gc l 1 2k km+- Cch 2: Gi x0 l khong cch t
v tr ( sao cho mt trong hai l xo khng b bin dng ) n VTCB ca vt m.
Gi s L02 c chiu di t nhin. Ta c+ Vt m VTCB : 0 1 0 20dh dhF F
+uuuur uuuuur. Chiu ln trc Ox, ta c: 2 0 1 0. .( ) 0 k x k d x (3).
Trong d = AB ( L01 + L02 ); x0 l khong cch t v tr m L02 khng b bin
dng n VTCB. + Xt vt m thi im t, c li x: 1 2.dh dhma F F +r uuur
uuuurChiu ln trc Ox: 2 0 1 0.( ) .( ) " k x x k d x x mx + (4).
Thay (3) vo (4) ta c 1 2 1 2" . . ( ) mx k x k x x k k +1 21 2" ( )
0 " . 0k kmx x k k x xm+ + + + . t2 1 2k km+ . Vy ta c: 2" . 0 x x
+ Cnghiml. ( ) x Acos t +. Vyvt mdaongiuhovi tnsgcl 1 2k km+ .*
Trng hp 3. AB
