5. SHM

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 95

SHM

EXERCISE # 1

3.6 Conserving momentum : 2V = 3V

V =3

2V..

Ei=

2

1m

1V

12 =

2

1.2.V2 = V2

2

1KA2 = V2.

Ef=

2

1.m

2V

22 =

2

1.3.

3

2.

3

2V2 =

3

2V2

2

1KA2 =

3

2V2 =

3

2E

i( Ei = V

2 from above)

2

1KA2 =

3

2(

2

1KA2) A =

3

2A Ans.

3.7

Rubber ribbon can exert only tension not compression so at a time only one is effective.

jcj fjcu dsoy ruko vkjksfir dj ldrk gS] lEihMu ugh vr% ,d le; esa dsoy ,d gh izHkkoh jgsxkA

T = 2K

m

3.8 =m

Kand vkSj Ke = mg (at mean position) (ek/; fLFkfr ij)

or;k =e

g

3.9 Just after cutting the string extension in spring =k

mg3

The extension in the spring when block is in mean position =k

mg

Amplitude of oscillation

A =k

mg3

k

mg=

k

mg2.

jLlh dkVus ds rqjUr ckn fLiax esa foLrkj =k

mg3

tc CykWd e/; fLFkfr ij gksrk gS rks fLizax esa foLrkj =k

mg

nksyu dk vk;ke

A =k

mg3

k

mg=

k

mg2.

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4.22mmV

2

1= 15 103

Vm

= s/m150.0

A = s/m150.0

L qm.

L

g= s/m150.0

gL = 310100

150.0

L =1.0

150.0= 1.5 m

4.3 With respect to the cart, equilibrium position of the pendulum is shown.

If displaced by small angle from this position, then it will execute SHM about this equilibrium position,time period of which is given by :

T = 2effg

L; g

eff= 22 )g3(g

geff

= 2g T = 1.0 second

5.2 I = 5

2

mR2 = 5

2

25 (0.2)2 = 5

2

= C c =

=1

1.0= 0.1

T = 2C

I= 2

1.05

2

= 2 2 = 4 secs

6.1 y = 4 cos2

2

tsin (1000 t)

y = 2cos

2t

2t1999sin

2t2001sin

y = sin (1001 t) + cos (1000 t) + sin (1000 t) + cos (1999 t)

y = sin (1001 t) + 2 sin (1000 t + /4) + cos (999 t)

So the given expression is composed by three equation of S.H.M.

vr% nh x;h lehdj.k rhu l-vk-x- dh lehdj.k ls cuh gSA

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EXERCISE # 2

PART-I

4. Acceleration Roj.k (A) = 2x amax

=

2

T

2

A

When amax

= g then block and piston will be separated

tc amax

= g oLrq rFkk fiLVu vyx gks tk;sxsA

amax

= g =

2

T

2

A ; (g = 2)

g =1

4 2A A =

4

1= 0.25 m

7. Let ekuk x1 = A1 sin 1t and x2 = A2 sin 2t

Two pendulums will v ibrate in same phase again when there phase difference (2 1)t = 2nks yksyd nqckjk leku dyk es dEiUu djsxs tc mudk dykUrj (2 1)t = 2gksxkA

(2T

2

1T

2)t = 2

1Tn44.1g

1

g

= 2 (where n is number of vibrations completed by longer pendulum)

(tgk n yEcs yksyd )kjk iwjs x;s nksyuks dh la[;k gS)

g44.1

2n44.1

g

1

g

= 2 n = 5

Thus after 5 vibrations of longer pendulum they will again start swinging in same phase.

vr% cMs yksyd ds 5 dEiUuks ds ckn ;s nqckjk leku dyk esa nksyu izkjEHk djsxsaA

8. a = 2x= 2 A sin t

=

T

0

T

0

2

dt

dt)t(sinA

=0T

tcosA

T

0

2

= 0

9. aavg

=

2/T

0

2/T

0

2

dt

dttsinA

2/T

tcosA

2/T

0

2

|a|avg

=2/T

]11[A 2

2

A4=

A2 2

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11. V = 22 xA(50)2 = (10)2 (102 x2)

x = 75 = 5 3

So, separation between points is

vr% d.kksa ds e/; dh nwjh x = 2 35 = 310 = 17.32 cm.

12. Consider SHM as projection of uniform circular motion.

From figure the phase difference between two particles is 120.

ljy vkorZ dks ,dleku oUrh; xfr dk iz{ksi ekurs gq,fp=k ls nks d.kks ds e/; dykUrj 120 gSA.

16. Let displacement of block is x1and of cart is x

2as shown

ekuk oLrq dk foLFkkiu x1

rFkk xkMh dk x2

gS fp=kkuqlkj

by linear momentum conservation

js[kh; laosx laj{k.k ls

mv1= Mv

2 v

2=

M

mv1so x

2=

M

mx1

For block Force equation can be written as

xqVds ds fy;s cy lehdj.k ,sls fy[kh tk ldrh gSAF = 2k(x

1+ x

2) = m2x

1

2k

11 x

M

mx = m2x

12 = 2k

Mm

mM

So vr% T = 2)mM(k2

Mm

17. Time period vkorZ dky T = 2k

where tgk =

21

21

MM

MM

= MM

MM

=

2

M

So, time period vr% vkorZ dky T =

2

M2 ( k = )

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18. T = 2g

, As it does not depend on amplitude pwafd ;s vk;ke ij fuHkZj ugh djrk gS

% change in time period is 0 % Hence option (D) is correct.vkoZr dky esa izfr'kr ifjorZu 0 % gS vr% fodYi (D) lgh gSA

19.

Let upper block is pushed down by x. at equilibrium mg = ky, i.e., weight of upper block is balanced by spring

when it is deformed by y. upper block will perform SHM with amplitude x about equilibrium position. lower block

will leave surface when spring is extended by y, means upper block is at distance 2y from its mean position.

That should be upper extreme position of upper block. So amplitude x = 2y

ekuk mPp CykWd uhps x /kdsyk tkrk gSA lkE;oLFkk ij mg = ky. mPp CykWd dk Hkkj fLizax }kjk larq fyr gksrk gS] tc fLizaxes ladqpu y gSA mPp CykWd lkE;oLFkk ds ifjr vk;kex ds lkFk l-vk-x- djsxkA tc fLizax y ls foLrkfjr gksrh gS] fuEu CykWdlrg dks NksM nsxk vFkkZr mPp CykWd bldh ek/;oLFkk ls 2y nwjh ij gksrk gSA ;g mPp CykWd dh mPp lhekUr fLFkfr gksuhpkfg;s vr% vk;kex = 2y.

Alternateat equilibrium of upper block mg = ky

lkE;oLFkk ij mPp CykWd ds fy;s mg = kylower plate will leave the surface if the extension in spring is y

;fn fLizax es foLrkj y gS rks fuEu IysV lrg NksM nsxhLet upper plate is displaced by x downward and left

ekuk mPp IysV x uhps foLFkkfir djrs gq;s NksMh tkrh gSso by energy conservation between compressed to extended positions

rks tkZ laj{k.k ls (laihfMr rFkk foLrkfjr voLFkkvksa ds e/;)

0 +2

1k (x + y)2 = mg (x + 2y) =

2

1ky2

21 kx2 +

21 ky2 + kxy = mgx + mg2y +

21 ky2

x = 2y

20. At equil ibrium :

mg sin = kA

A =K

sinmg

and time period for spring block system is :

T =k

m2

Hence (A)

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21.

Force on fluid nzO; ij cyF = sg (x + xcos)= m2x m2 = sg (1 + cos)

= m

)cos1(sg

T =g)cos1(s

m2

23. For simple pendulum = /g and maximum linear displacement x0 = and equation of S.H.M

ljy yksyd ds fy;s = /g vkSj vf/kdre js[kh; foLFkkiu x0 = vkSj l-vk-x- dh lehdj.k

x = x0

cos t

x = cos tg

24. Point O is moving as shown fcUnq O fp=kkuqlkj xfr dj jgk gSASo acc. Of particle w.r.t O vr% d.k dk O ds lkis{k Roj.k

= j)g(i( 2

So vr% geff

. = 222

1 )g(

So time period vr% vkorZdky

= 22

1

))g(( 222

1

26. x = 10 sin3(t) ;

4

)A3sin(Asin3Asin3

x = 10

4

)t3sin()tsin(3

x =4

30sin t

4

10sin 3t

So Amplitude vr% vk;ke=4

30,

4

10

frequency vko fk = 1/2 , 3/2

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28. At time t1, veloci ty of the particle is negative i.e. going towards Xm. From the graph, at time t1, its speed

is decreasing. Therefore particle l ies in between Xm and 0.

At time t2, velocity is positive and its magnitude is less than maximum i.e. it has yet not crossed O.

It lies in between Xm and 0.

Phase of particle at time t1 is (180 + 1).Phase of particle at time t2 is (270 + 2)Phase difference is 90 + (2 1)2 1 can be negative making < 90 but can not be more than 90.

t1le; ij d.k dk osx _.kkRed gS vFkkZrXmdh rjQ xfr'khy gSA xzkQ ls t1le; ij bldh pky ?kV jgh gSA blfy,

d.k XmrFkk 0 ds e/; fLFkr gSA

t2le; ij] osx /kukRed gSA rFkk bldk ifjek.k vf/kdre ls de gSA vFkkZr~ ;g O dks ikj ugh djrk gSA ;gXmrFkk

0 ds e/; fLFkr gSA

t1le; ij d.k dh dyk (180 + 1).

t2le; ij d.k dh dyk (270 + 2)

dykUrj 90 + (2 1)

< 90 ds fy, 2 1_.kkRed gks ldrk gSA ijUrq 90 ls T;knk ugh gks ldrk gSA

30. maxm amplitude vf/kdre vk;ke= So time to travel angle vr% dks.k r; djus esa yxk le;

x = x0

sin wt

= sin t

t =1

sin1

Now period of pendulum =2

1(actual time period) +2 (time to travel angle )

vc yksyd dk vkoZrdky =2

1(okLrfod vkorZdky) +2 (dks.k r; djus es yxk le;)

=2

1. 2

g

+ 2

g

sin1

= 2g

1sin2/

Also if time taken to travel from right extreme position to wall is t then

;fn nka;h lhekUr fLFkfr ls nhokj rd tkus esa yxk le; t gS rks = cos t

cost =

t =

cos1 1

T = 2t =

cos

2 1

T =

cosg

2 1

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PART-II

1. The cylinder moves a distance L to left in time interval t = 0 sec to t =4

T=

2

k

m. The mass m remains

at rest in that time interval.

The speed of cylinder at t =2

k

mis

vmax

= L =m

kL.

At t =2

k

mthe mass m collides elastic with the right wall of cylinder of mass m. The cylinder comes to

rest and the mass moves further left with speed vmax

. After time t =maxv

Lthe mass m collides with left side

of cylinder. The mass comes to rest due to elastic collision and the cylinder further moves to left for time

2

k

m. This motion is one half of oscillation.

Time period of oscillation is

2

k

m

2v

L

k

m

2 max= 2

k

m[ + 1] Ans.

2. Frequency (vko fk) n2

=21 mm

k

2

1

=54

100

2

1

=3

10

2

1

=

35

By conservation of l inear momentum at mean position,

ek/; voLFkk ij js[kh; laosx laj{k.k ls

Pi = P f m1

1A

1= (m

1+ m

2)

2A

2

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11

1 Am

km = (m

1+ m

2) 2

21

Amm

k

11Akm = 221 A)mm(k

A2 = 302 m.

3. (A) Initial position of block is an extreme position. CykWd dh izkjfEHkd voLFkk ,d lhekUr fLFkfr gSAt equilibrium lkE;oLFkk ij F = KA A = F/KTime periodvkorZdky

T = 2K

m

(B) Kinetic energy at mean position ek/; voLFkk ij xfrt tkZ

K.E. =

2

1K AA2

=2

1K

2

K

F

=

K2

F2

(C) Total energy of the block. CykWd dh dqy tkZ

2

1K AA2 =

2

1K

2

K

F2

=

K2

F2

4. T1= Time period for Ist half T

1= 2

g

= 2g

2.1

T1

= igys Hkkx dk vkorZdky T1

= 2g

= 2

g

2.1

T2= Time period for IIst half T

2= 2

g

= 2

g

1

T2= nwljs Hkkx dk vkorZdky

So total time taken in one oscillation

vr% ,d nksyu esa fy;k x;k dqy le;

T =2

T1 +2

T2

=g

1

2

2

g

2.1

2

2

= 2.1 sec

5.

At t = 0 Particle 2 is at point B and moving towards origin so displacement

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At t = 0 ij d.k 2 fcUnq B ij gS rFkk ewy fcUnq dh rjQ tk jgk gS vr%Y = 4 A sin tY = 4 2 sin tand displacement of particle 1 is X = 3 A cos tvkSj d.k 1 dk foLFkkiu gS X = 3 A cos t X = 3 2 cos t

So distance between them vr% vuds e/; nwjh = 22 YX d2 = 29 (16 sin t + 12 cos t) = 29 4 (4 sint + 3 cost) 29 20 (sint + 37)

So vr% 2maxd = 49 dmax = 72mind = 9 dmin = 3

6. (i)

force on particle at point P

fcUnq P ij fLFkr d.k ij cy

F =a

)xa(mg2

a

)x(mg

F =a

mg(2a 3x)

F =a

mg3(x 2a/3)

(ii) So this is equation of S.H.M (F = mx) so particlevr% ;s lehdj.k l-vk-x- dh lehdj.k gSA (F = mx) vr% d.k l-vk-x-Perform S.H.M with mean position x 2a/3

= 0

djsxk ftldh ek/; voLFkk gS x 2a/3

= 0 x = 2a/3

(from point A) fcUnq A ls

So vr% 2 =a

g3 = a/g3

So Time period vr% vkoZrdky T = 2 g3/a

and amplitude vkSj vk;ke= 2a/3 a/2 = a/6(iii) minimum distance from B ls U;wure nwjh

= a (2a/3 + a/6) = a/6

(iv)

at point P velocity of particle = 0

fcUnq P ij d.k dk osx = 0

and force vkSj cy =a

xmg2

(at point q) (q fcUnq ij )

acc. Roj.k =a

xg2

v

o

dvv = 6/a

oa

dxgx2 V =

6

ga2

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EXERCISE # 3

1.2 (A) In frame of lift effective acceleration due to gravity is2

g3

2

gg downwards

fyV ds se esa xq:Ro ds dkj.k Roj.k =

2

g3

2

gg uhps dh rjQ

T = 2g3

2

(B) K = mg L

g

m

k

constant acceleration of lift has no effect in time period of oscillation.

fyV dk fu;r Roj.k] nksyu ds vkorZdky ij dksbZ vlj ugha Mkyrk gSA

T = 2k

m= 2

g

(C) T = 2mgd

= 2 g3

22

2mg

3m

2

(D) T = 2Ag

m

= 2 AgA2/

= 2g2

2.4 When speed of block is maximum, net force on block is zero. Hence at that instant spring exerts a force

of magnitude 'mg' on block.

tc CykWd dh pky vf/kdre gS CykWd ij ifj.kkeh cy 'kwU; gS vr% ml {k.k fLizax CykWd ij mg ifjek.k dk cy yxk;sxkA

2.5 At the instant block is in equilibrium posi tion, i ts speed is maximum and compression in spring is x given

by kx = mg .... (A)

From conservation of energy

CykWd ds lkE;oLFkk ds {k.k] CykWd dh pky vf/kdre gksxh rFkk ml {k.k fLizax esa laihMu x ds fy,kx = mg .... (A)

tkZ laj{k.k ls

mg (L + x) =2

1kx2 +

2

1mv 2

max.... (B)

from (A) and (B) we get vmax

= gL2

3.

(A) o (B) ls vmax

= gL2

3.

2.6 Vmax

= gL2

3and =

m

k= 2

L

g

A =maxV

=4

3L

Hence time taken t, f rom start of compression till block reaches mean position is given by

blfy;s laihMu ds izkjEHk ls CykWd }kjk ek/; fLFkfr rd igqpus esa fy;k x;k le; t fn;k tk ldrk gS]

x = A sin t0 where tgk x = 4L

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t0

=g4

Lsin1

3

1

Time taken by block to reach from mean position to bottom most position is

4

2=

g

L

4

Hence the required time =g

L

4

+

g4

Lsin1

3

1

CykWd }kjk ek/; fLFkfr ls lcls fupyh fLFkfr rd igqpus es a fy;k x;k le;

4

2=

g

L

4

blfy;s vko';d le; =g

L

4

+

g4

Lsin1

3

1

4.2 In equilibrium lkE;koLFkk ds fy, mg = k x

=m

k=

x

gand vkSj T =

2

4.3 x2

= 5 2 2 sin (2t + ) = 10 sin (2t + ) where is tan1 (A) = 45 ;gk, tan1 (A) = 45 gSA

so ratio is 1 : 2

vr% vuqikr 1 : 2 gSA

EXERCISE # 4

1. U (x) = )e1(K2

xIt is an exponentially increasing graph of potential energy (U) with x2 . Therefore U versus x graph will be

as shown.

From the graph it is clear that at origin___

Potential energy U is minimum ( therefore, kinetic energy will be maximum ) and force acting on the

particle is also zero because F =dx

dU= (slope of U x graph ) = 0.

Therefore, origin is the stable equilibrium position. Hence particle will oscillate simple harmonically about

x = 0 for small displacements. Therefore, correct option is (D).

(A) At equilibrium position F =dx

dU= 0 i.e. slope of U-X graph should be zero and from the graph we can

see that slope is zero at x = 0 and x = . Now among these equilibriums stable equilibrium position isthat where U is minimum ( Here x = 0 ). Unstable equilibrium position is that where U is maximum ( Here

none). Neutral equilibrium position is that where U is constant ( Here x = ). Therefore, option (A) iswrong.

(B) For any finite non-zero value of x, force is directed towards the origin,

because origin is in stable equilibrium position. Therefore, option (B) is

incorrect.

(C) At origin, potential energy is minimum, hence kinetic energy will be

maximum. Therefore, option (C) is also wrong.

U

K

XU (x) = )e1(K

2x

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;g fLFkfrt tkZ (U) dk x2 ds lkFk pj?kkrkadh c

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)223()12(a

A

E

E 22

dy,

ifj.kkeh

Eifj.kkeh

= (3 + 22 ) E,dy

3. Free body diagram of bob of the pendulum with respect to the accelerating frame of reference is as fol lows:

Net tension in the string is T = mg cos

So, geff

=m

T=

m

cosmg = g cos

T = 2effg

Lor T = 2

cosgL

Alternative :

Whenever point of suspension is accelerating

Take T = 2effg

LWhere a.gg

eff

a

= Acceleration of point of suspension.

In this question a

= g sin a (down the plane)

ag = g

eff= )(90cos)sin(g2(g))sing(g 22 = g cos

Rofjr funsZ'k ra=k ds lkis{k isUMwye ds yksyd dk eqDr oLrq js[kk fp=k ,slk gksxkA jLlh esa dqy ruko T = mg cos

vkSj yksyd dk dqy Roj.k geff

=m

T=

m

cosmg = g cos

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T = 2effg

L;k T = 2

cosgL

oSdfYid

tc Hkh vkyEcu fcUnq Rofjr gksrk gSA

yhft;ss T = 2effg

Ltgk a.gg

eff

a

= vkyEcu fcUnq dk Roj.k

bl iz'ku esa a= g sin (ry ds uhps dh vksj)

ag = g

eff= )(90cos)sin(g2(g))sing(g 22 = g cos

4. In SHM , velocity of particle also oscillates simple harmonically. Speed is more near the mean position

and less near the extreme positions.

Therefore, the time taken for the particle to go from 0 to A/2 will be less than the time taken to go from A/

2 to A. or T1

< T2

.

SHM esa d.k dk osx ljy vkorZ xfr ls nksfyr gksrk gSA ek/; fLFkfr ds fudV pky vf/kd o vk;ke fLFkfr ds ikl ;g de

gksrh gSA

vr% d.k }kjk 0 ls A/2 rd tkus esa yxk le; A/2 ls A rd tkus esa yxs le; ls de gksxk] vFkkZr~ T1

< T2

5. Potential energy is minimum (in this case zero) at mean position (x = 0) and maximum at extreme positions (x

= A)

At time t = 0, x = A. Hence P.E. should be maximum. Therefore graph is correct.Further in graph III, P.E. is minimum at x = 0. Hence this is also correct.

ek/; fLFkfr (x = 0) ij fLFkfrt tkZ U;wure gS bl fLFkfr esa 'kwU; gS vkSj pje fLFkfr (x = A) ij vf/kdre gSAt = 0, x = A vr% fLFkfrt tkZ vf/kdre gksuh pkfg,] blfy, vkjs[k lgh gSAnqckjk vkjs[k III esa, fLFkfrt tkZ x = 0 ij U;wure gSA vr% ;g Hkh lgh gSA

6. Half of the volume of sphere is submerged.

For equilibrium of sphere,

weight = upthrust

V sg =

2

V(

L) (g)

s=

2

L

When slightly pushed down by x weight will remain as it is while upthrust will increase. The increased upthrust

will become the net restoring force (upwards).

F = (extra upthrust)= (extra volume immersed) (

L) (g)

or ma = (R2) xLg (a = acceleration)

3

4R3

2L

a = (R2Lg) x a =

R2

g3x

as a x motion is simple harmonicFrequency of oscillation,

f =x

a

2

1

= R2g3

2

1

.

xksys dk vk/kk vk;ru Mwck gqvk gS

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xksys dh lkE;koLFkk ds fy,Hkkj = mRIykod cy

V sg =

2

V(

L) (g)

s=

2

L

tc FkksM+k lk uhps dh vksj x ls /kdsyk tkrk gS Hkkj vifjofrZr jgrk ijUrq mRIykod cy c

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9. x = A2

)t2cos1( + B

2

)t2cos1( +

2

Csin 2t

Choose different combinations of A, B & C to get linear combination of sin & cos functions.

A, B rFkk C ds fHkUu&fHkUu ekuksa ds fy;s sin o cos Qyu ds js[kh; feJ.k izkIr djksA

10. (A) It is similar equation as v = 22 xa in SHM.

(B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing.

(C) It is spring mass system performing SHM.

(D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocity so

it will never return back.

(A) ;s l-vk-x- esa lehdj.k v = 22 xa ds leku gSA

(B) tSls- tSls x ?kVrk gS d.k ?kVrh gqbZ pky ls /kukRed x-v{k ij ewy fcUnq dh xfr djrk gSA

(C) ;g l-vk-x.djrk gqvk fLizax nzO;eku fudk; gSA

(D) oLrq iFoh ls nwj tk jgh gS vr% bldh pky ?kVsxh] pwfd bldh pky iyk;u osx ls T;knk gS vr% okil ugh yksVsxkA11. (A) From nature of SHM, the graph of potential energy as function of displacement will be parabolic

graph as given in option p

Hence (A) (p)(B) a = 0 or a = constant. (as per given condition)

V > 0 moving along positive x-axis

y displacement

y = ut 2

1at2 for a = constant

y = vt for a = 0

These two conditions are satisfied by (q) and (s).

(B) (q, s)(p) Is rejected because at t = 0 the displacement is not zero and velocity has negative values.

(C) R =g

)2sin(u2 and R u2 for a fixed angle of projection.

at u = 0, R = 0

(C) (s)

(D) T =g

2

T2 =g

4 2

y = xg

4 2

(D) (q)

(A) SHM dh izdfr ls] fodYi p ls fn;s x, xzkQ ls fLFkfrt tkZ&foLFkkiu dk Qyu ijoy; gksxkvr% (A) (p)

(B) a = 0 ;k a = fu;r (nh xbZ 'krZ ls)V > 0;g /kukRed x-v{k ds vuqfn'k gSAy foLFkkiu

y = ut 2

1at2 for a = fu;r

y = vt for a = 0

;g nksuksa fLFkfr;ka (q) o (s) }kjk larq"V gksrh gSA(B) (q, s)

(p) dks ugh fy;k x;k gS pwafd t = 0 ij foLFkkiu 'kwU; ugh gS rFkk osx dk _.kkRed eku gSA

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(C) R =g

)2sin(u2 and R u2 ,d fu;r iz{ksi.k dks.k ds fy,

rFkk u = 0, R = 0(C) (s)

(D) T =g

2

T2 =g

4 2

y = xg

4 2

(D) (q)

12. From graph (xzkQ ls)

T = 8 second., A = 1 cm, x = A sint. = 1sin8

2t.

a = 2x = 2

8

2

sin

8

2t cm/s2

At, t =3

4second ij a =

2

8

2

sin

3

=

32

3 2cm/s2.

13.

Torque about P = (kx)2

L+ (kx)

2

L= kxL = k

2

L2 ( x =

2

Lsin ~ 2

L )

=

12

ML

2

KL 22

M

K6 = =

M

K6 = 2

=M

K6and f =

2

=M

K6

2

1

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P ds lkis{k cy vk?kw.kZ = (kx)2

L+ (kx)

2

L= kxL = k

2

L2 ( x =

2

Lsin ~ 2

L )

=

12

ML2

KL22

MK6 = = M

K6 = 2

=M

k6vkSj f =

2

=M

k6

2

1

14.

Extensions in springs are x1

and x2

then

fLaxksa esa foLrkj x1o x

2gS rc

k1x

1= k

2x

2

andvkSj x1

+ x2

= A

x1

+2

11

k

xk= AA x

1=

21

2

kk

Ak

15. When 0 < E < V0 there will be acting a restoring force to perform oscillation because in this case particle will be

in the region |x| x0 .tc 0 < E < V0 , ;gk nksyu djus ds fy, ,d izR;ku;u cy dk;Zjr gS D;ksfd bl fLFkfr esa d.k {ks=k |x| x0esa jgrk gSA

16. V = x4

T.E. =2

1m2A2 = A4 (not strictly applicable just for dimension matching it is used)

2 =m

A2 2 T

m

A

1

V = x4

T.E. =2

1m2A2 = A4 (iw.kZr% ykxw ugh gS dsoy ;g foeh; :i ls feyku ds fy, mi;ksxh gS)

2 =m

A2 2 T

m

A

1

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17. F =dx

dU

as for |x| > x0 V = V0 = constant

tSlk fd |x| > x0ds fy, V = V0 = fu;rkad

dx

dU= 0

F = 0.

18. n = mk

=m

/yA =

m

yA

1.01

)109.4()10n( 79

= 140 n = 4.

19.

Restoring torque in both cases will be equal but moment of inertia of rotating system will be less in case B as

disc is free to rotate about its centre, so time period will be less for case B and hence angular velocity will be

greater.

20. For a ball thrown upward momentum is +ve max at the intial position (Postion zero)

And then when it returns to the same position zero its momentum gets inverted (ie becomes negative)

21. We have

1

2m2 AA2 = E

E1=

1

2m2(2a) 2 E

2=

1

2m2 a2

E1

= 4 E2

22. Momentum is zero as the block starts from rest when the spring is compressed to its maximum (position of

block will be taken as positive by the convention given) and then it comes down (momentum negative) ; the

block moves down till the spring is compressed again to its maximum (this time the position is negative) and

from this position the block starts to move up(momentum positive). moreover there is loss of energy in process

Final momentum is less than initial momentum

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23.

B = A

=2

3

+

2

3

=

4

3

24. x

X = A cos t

x = 0.2 cos 13

x = 0.1

Horizontal distance travelled by stone = 5 m

5 =2

V 1

V = 5 2 = 50

PART - II1. Kinetic energy of particle of mass m in SHM at any point is

ljy vkorZ xfr esa m nzO;eku dh fdlh Hkh fcUnq ij xfrt tkZ= m2 (a2 x2)

and potential energy rFkk fLFkfrt tkZ =

22xm2

where a is amplitude of particle and x is the distance from mean position.

tgk a d.k dk vk;ke gS rFkk x ek/; voLFkk ls nwjh gSAAt mean position, x = 0

vr%] ek/; voLFkk ij] x = 0

So, K.E. =22am

2

(maximum vf/kdre)

P.E. = 0 (minimum U;wure)

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2. T = 2k

M......(i)

T = 2k

mM

3

T5= 2

k

mM ......(ii)

Form equation (i) and (ii)

lehdj.k (i) rFkk (ii)ls

5

3=

mM

M

mM

M

25

9

9M + 9m = 25M

16M = 9m

9

16

M

m

3. T = 2g

l

2

1d

2

1

T

dT

l

l 21% = 10.5%

4. Amplitudevk;ke

a = 22 44 = 4 2

5. At x = 0, kinetic energy is maximum and potential energy is minimum.

x = 0 ij xfrt mtkZ vf/kdre gS rFkk fLFkfrt mtkZ U;wure gSA

6. The time period of simple pendulum in air

ok;q esa ljy yksyd dk vkorZ dky

T = t0 = 2

g

........... (i)

, being the length of simple pendulum.

, ljy yksyd dh yEckbZ gSaSAIn water, effective weight of bob

ty esa xksyd dk izHkkoh Hkkjw = weight of bob in air upthrust

w = ok;q esa xksyd dk Hkkj ij dh vksj cy Vg

eff= mg mg

=Vg Vg = ( )Vg

where = density of bob,

tgk =xksyd dk ?kuRo = density of waterty dk ?kuRo

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geff

=

'g =

'1 g

t = 2

g

'

1

(ii)

Thus vr%,0t

t=

'

1

1

=

10003/41000

1

1= 2

t = 2 t0

7. Time period of spring fLizax dk vkorZdky

T = 2

k

m

k, being the force constant of spring.

k, fLizax dk cy fu;rkad gSAFor first spring igyh fLizax ds fy,

t1

= 2

1k

m................(i)

For second spring nwljh fLizax ds fy,

t2

= 2

2k

m

The effective force constant in their series combination is

Js.kh e la;kstu dk izHkkoh cy fu;rkad

k =21

21

kk

kk

Time period of combination la;kstu dk vkorZ dky

T = 2

21

21

kk

kkm

T2 =

21

212

kk

kkm4 ..............(iii)

From equations (i) and (ii), we obtain

lehdj.k (i) o (ii) ls

t21

+ t22= 42

21 k

m

k

m

t21

+ t22= 42 m

21 k

1

k

1

t21

+ t22=

21

212

kk

kkm4

t21+ t2

2= T2 [from equation (iii)] lehdj.k (iii)ls

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8. In simple harmonic motion when a particle is displaced to a position from its mean position, then its kinetic

energy gets converted into potential energy and vice-versa. Hence, total energy of a particle remains constant or

the total energy in simple harmonic motion does not depend on displacement x.

ljy vkorZ xfr esa tc d.k dks ek/; voLFkk ls fdlh fLFkfr ij foLFkkfir djrs gS rks bldh xfrt tkZ fLFkfrt tkZ esa cnytkrh gS rFkk bldk foijhr Hkh lR; gSA vr% d.k dh dqy tkZ fu;r jgrh gS ;k d.k dh dqy tkZ x ls Lora=k gksrh gSA

9. nitial angular velocity of particle = 0

and at any instant t, angular velocity = Therefore, for a displacement x, the resultant acceleration

f = (20 2) x ................(i)

External force F = m (202) x ................(ii)

Since, F cos t (given) From eq. (ii)m (2

02) x cos t ................(iii)

Now, equation of simple harmonic motion

x = A sin (t + ) ................(iv)at t = 0 ; x = A

A = A sin (0 + )

=2

x = A sin

2t = A cos t ................(v)

Hence, from equation (iii) and (v), we finally get

m (20 2) A cos t cos t

A 220m1

d.k dk izkjfEHkd dks.kh; osx = 0

rFkk fdlh le; t ij d.k dk dks.kh; osx = vr% x foLFkkiu ds fy,] ifj.kkeh Roj.k

f = (20 2) x ................(i)

ck cy F = m (202) ................(ii)

pwafd, F cos t (fn;k gS) lehdj.k (ii)ls

m (202) x cos t ................(iii)

vc, ljy vkorZ xfr dh lehdj.k lsx = A sin (t + ) ................(iv)t = 0 ij ; x = A

A = A sin (0 + )

=2

x = A sin

2t = A cos t ................(v)

vr%, lehdj.k (iii) rFkk (v) ls,m (2

0 2) A cos t cos t

A 220m1

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10. For amplitude of oscillation and energy to be maximum, f requency of force must be equal to the initial frequencyand this is only In resonance state

1=

2

nksyu ds vk;ke rFkk mtkZ ds vf/kdre gksus ds fy,] cy dh vkofk izkjfEHkd vkofk ds cjkcj gksuh pkfg, rFkk ;g dsoy vuqukndh voLFkk

1=

2esa gh gksrk gSA

11. 2

2

dtxd = x ........ (i)

We know ge tkurs gS fd a = 2

2

dt

xd= 2x ........ (ii)

From Eq. (i) and (ii), we have

lehdj.k (i) o (ii) ls]2 =

=

or;kT

2= T =

a

2

12. T < T1> (T

2= T)

G

l

Spherical hollow ballfilled with water

l + l

Spherical hollow ballhalf filled with water

G

G

T = 2g

lT

1

= 2g

ll

G

l

Spherical hollow ball

T2

= 2g

l= T.. and rFkk T

1> T

2

Hence, time period first increases and then decreases to the original value.vr%] vkorZ dky igys c

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Rise Academy 607/608 A Talwandi Kota (Raj ) 324005 120

16. x = x0cos

4t , v = x

0sin

4t

a = x02

4tcos , a = x

02

4tcos

a = x02

4

3tcos .

17.m

kk

2

1 21

,

m

k4k4

2

1 21new

= 2.

18. Kav

=

4/T

0

4/T

0

2

dt

dt)]tcos(a[m2

1

= 4/T

0

222

dt)t(cos

4

T.2

ma

=8

T.

T

ma2 22=

4

1ma22

=4

1ma2(2)2 = 2 ma2 2.

19. x = A sin tv = A cos ta = A2 sin t

(1)x

aT=

tsinA

tsinA 2

T = 2T = 2

(2) aT + 2 = A2

T sin t + 2. A cos t

(3)V

aT=

tcosA

TtsinA 2

(4) a2T2 + 422 = + A24 T2 sin2t + 42A22 cos2t = + A242

2

sin2t + 42A22 cos2t = 42A22

20. At mean position, the total force is zero so

dt

PdF

= 0 P = constant

P = m1v1 = m2v2

M

KM = (M + m)

mM

KA2

2

1

A

A

M

mM

21. x1

= A sin tx

2= X

0+ A sin (t + )

x2 x

1= X

0+A[sin (t + )sint]

=X0 + 2A sin /2) comparing X0

+ 2 A sin (f/2) with X0

+ we get f = f/3

5. SHM

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