Simple Harmonic Motion

Chapter # 12 Simple Harmonic Motion

Page # 1manishkumarphysics.in

SOLVED EXAMPLES1. The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away from

the centre of oscillation. Find the spring constant.Sol. The simple harmonic motion is defined as

F = k x.

The spring constant is k = xF

= cm5N4

= m105N4

2= 80 N/m.

2. A particle of mass 0.50 kg executes a simple harmonic motion under a force F = (50 N/m)x. If it crosses thecentre of oscillation with a speed of 10 m/s, find the amplitude of the motion.

Sol. The kinetic energy of the particle when it is at the centre of oscillation is

E =21

mv2

=21

(0.50 kg) (10 m/s)2

= 25 J.The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the

kinetic energy is zero. The potential energy here is21

kA2. As there is no loss of energy,,

21

kA2 = 25 J .............(i)

The force on the particle is given byF = (50 N/m)x.

Thus, the spring constant is k = 50 N/m.Equation (i) gives

21

(50 N/m) AA2 = 25 J

or, A = 1 m.

3. A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring ofspring cosntant 80 N/m. Find the time period.

Sol. The time period is

T = 2km

= 2 m/N80kg10200 3

= 2 0.05 s = 0.31 s.

4. A particle executes simple harmonic motion of amplitude A along the X-axis. At t = 0, the position of theparticle is x = A/2 and it moves along the positive x - direction. Find the phase constant if the equation iswritten as x = A sin (t + ).

Sol. We have x = A sin (t = ). At t = 0, x = A/2.Thus, A/2 = A sinor, sin = 1/2or, = /6 or 5/6.

The velocity is v = dtdx

= A cos(t + ).



At t = 0, v = A cos.

Now cos 6

=23

and cos 65

= 23

As v is positive at t = 0, must be equal to /6.

5. A particle of mass 40 g execute a simple harmonic motion of amplitude 2.0 m. If the time period is 0.20 s, findthe total mechanical energy of the system.

Sol. The total mechanical energy of the system is

E =21

m 2 AA2 =21

m2

T2

A2 = 2

22

TAm2

= 22232

)s20.0()m100.2)(kg1040(2

= 7.9 103 J

6. A body makes angular simple harmonic motion of amplitude /10 rad and time period0.05 s. If the body is ata displacement /10 rad at t = 0, write the equation giving the angular displacement as a function of time.

Sol. Let the required equation be = 0 sin (t + )

Here 0 = amplitude = 10

rad

=T2

= s05.02

= 40 s1

so that =

rad10 sin ts40 1 ..............(i)

At t = 0, = /10 rad. Putting in (i)

10

=

10 sin

or, sin =1or, = /2Thus by (i),

=

rad10 sin

2ts40 1 =

rad10 cos [(40 s

1) t ].

7. Calculate the time period of a simple pendulum of length one meter. The acceleration due to gravity at theplace is 2 m/s2.

Sol. The time period is

T = 2 g/ = 222 s/m

m00.1

= 2.0 s.

8. In a laboratory experiment with simple pendulum it was found that it took 36 s to complete 20 oscillationswhen the effective length was kept at 80 cm. Calculate the acceleration due to gravity from these data.

Sol. The time period of a simple pendulum is given by

T = 2 g/

or, g = 22

T4

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

In the experiment described in the question, the time period is

T = 20s36

= 1.8 s.

thus, by (i)



g = 22

)s8.1(80.04

= 9.75 m/s2.

9. A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitudeunder gravity. Find the time period of oscillation.

Sol. For small amplitude the angular motion is nearly simple harmonic and the time period is given by

T = 2 mg

= 2

mg)3/m( 2

= 2 g3

= 2 2s/m80.93m00.1

= 1.16 s.

10. A uniform disc of radius 5.0 cm and mass 200 g is fixed at its centre to a metal wire, the other end of whichis fixed with a clamp. The hanging disc is rotated about the wire through an angle and is released. If the discmakes torsional oscillations with time period 0.20 s, find the torsional constant of the wire.

Sol. The situation is shown in figure. The moment of inertia of the disc about the wire is

=2

mr2 =2

)m100.5)(kg200.0( 22

= 2.5 104 kg - m2.The time period is given by

T = 2k

or, k = 22

T4

= 2242

)s20.0()mkg105.2(4

= 0.25 22

smkg

.

11. Find the amplitude of the simple harmonic motion obtained by combining the motionsx1 = (2.0 cm) sint

and x2 = (2.0 cm) sin (t + /3).

Sol. The two equations given represent simple harmonic motions along X-axis with amplitudesA1 = 2.0 cm andA2= 2.0 cm. The phase differnce between the two simple harmonic motions is /3. The resultant simpleharmonic motion will have an amplitude A given by

A = cosAA2AA 2122

21

= 3cos)cm0.2(2)cm0.2()cm0.2( 222 = 3.5 cm

QUESTIONS FOR SHORT ANSWER1. A person goes to bed at sharp 10.00 pm every day. It is an example of periodic motion? If yes, what is the

time period ? If no, why ?

2. A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force onthe particle zero at these positions according to Newtons first law?

3. Can simple harmonic motion take place in a noninertial frame? If yes, should the ratio of the force applied



with the displacement be constant ?

4. A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can yousay what is its displacement ? If you are told that its velocity at this instant is maximum, can you say whatis its displacement ?

5. A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed bythe sun on a horizontal plane move in a simple harmonic motion ?

6. A particle executes simple harmonic motion. Let P be a point near the mean position and Q be a point nearan extreme. The speed of the particle at P is larger then the speed at Q. Still the particle crosses P and Qequal number of times in a given time interval. Does it make you unhappy ?

7. In measuring time period of a pendulum, it is advised to measure the time between consecutive passagethrough the mean position in the same direction. This is said to result in better accuracy than measuringtime between consecutive passage through an extreme position. Explain.

8. It is proposed to move a particle in simple harmonic motion on a rough horiozntal surface by applying anexternal force along the line of motion. Sketch the graph of the applied forceagainst the position of theparticle. Note that the applied force has two values for a given position depending on whether the particle ismoving in positive or negative direction.

9. Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potentialenergy at some point other than the mean position ?

10. The energy of a system in simple harmonic motion is given by E =21

m2A2. Which of the following two

statements is more appropriate ?(A) The energy is increased because the amplitude is increased.(B) The amplitude is increased because the energy is increased.

11. A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?

12. Can a pendulum clock be used in an earth satellite ?

13. A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simplependulum is valid with the distance between the point of suspension and centre of mass of the bob acting asthe effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary ?

14. A block of known mass is suspended from a fixed support through a light spring. Can you find the time periodof vertial oscillation only by measuring the extension of the spring when the block is in equilibrium.

15. A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band isstopped and the soldiers are ordered to break the steps while crossing a bridge. Why ?

16. The force acting on a particle moving along X-axis is F = k(x v0t) where k is a positive cosntant. Anobserver moving at a constant velocity v0 along the X-axis looks at the particle. What kind of motion does hefind for the particle?

Objective - I1. A student says that he had applied a force F k/x on a particle and the particle moved in simple harmonic

motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x andno other force acted on the particle.,d fo|kFkhZ dgrk gS fd mlus fdlh d.k ij cy F k/xyxk;k gS rFkk d.k ljy vkorZ xfr dj jgk gSA og ;gcrkus ls euk dj jgk gS fd k fu;r gS ;k ugha A ;fn ;g eku fy;k tk;s fd mldk dk;Z dsoy /kukRed x fn'kkds fy;s gh gS rFkk d.k ij vU; dksbZ cy dk;Zjr ugha gS &(A*) As x increases k increases x c



(C) As x increases k remains constant. x c



fp=k esa nks ljy vkorZ xfr;k fn[kk;h x;h gSA nksuks a xfr;ksa esa fdu izkpyksa ds vyx&vyx eku gksaxsA

(A) amplitude (B) frequency (C*) phase (D) maximum velocity(A)vk;ke (B)vko`fk (C*) dyk (D)vf/kdre osx

11. The total mechanical energy of a spring - mass system in simple harmonic motion is

E =21

m 2A2. Suppose the oscillating particle is replaced by another particle of double the mass while the

amplitude A remains the same. The new mechanical energy will

ljy vkorhZ xfr esa fLizax nzO;eku fudk; dh dqy ;kaf=kd tkZ E =21

m 2A2 gSA eku fyft;s fd nksyu djus okys

d.k dks nqxus nzO;eku okys d.k ls izfrLFkkfir dj fn;k tkrk gS tcfd vk;ke AvifjofrZr j[kk tkrk gSA u;h ;kaf=kdtkZ &(A) become 2 E (B) become E/2 (C) become 2 E (D*) remain E(A) 2 E gks tk;sxh (B) E/2 gks tk;sxh (C) 2 E gks tk;sxh (D*) E gh jgsxhA

12. The average energy in one time period in simple harmonic motion isljy vkorhZ xfr esa ,d vkorZdky esa vkSlr tkZ gksrh gS &

(A*)21

m 2 AA2 (B)41

m 2 AA2 (C) m 2 A2 (D) zero 'kwU;

13. A particle executes simple harmonic motion with a frequency v. The frequency with which the kinetic energyoscillates is,d d.k vvko`fr ds lkFk ljy vkorhZ xfr dj jgk gSA xfrt tkZ ftl vko`fr ls nksyu dj jgh gS] og gS&

(A) v/2 (B) v (C*) 2v (D) zero 'kwU;

14. A particle executes simple harmonic motion under the restoring force provided by a spring. The time periodis T. If the spring is divided in two equal parts and one part is used to continue the simple harmonic motion,the time period will(A) remain T (B) becomes 2T (C) become T/2 (D*) become T/ 2,d fLizax }kjk yxk;s x;s izR;ku;u cy ds izHkko esa ,d d.k ljy vkorhZ xfr dj jgk gSa vkorZdky T gSA ;fn fLizaxdks nks cjkcj Hkkxksa esa foHkkftr djds] ,d Hkkx dk mi;ksx ljy vkorhZ xfr ds fy;s fd;k tkrk gS] rks vkorZdky&(A) T jgsxk (B) 2T gks tk;sxk (C) T/2 gks tk;sxk (D*) T/ 2 gks tk;sxk

15. Two bodies Aand B of equal mass are suspended from two separate massless springs of spring cosntant k1and k2 respectively. If the bodies socillate vertically such that their maximum velocities are equal, the ratio ofthe amplitude of A to that of B is HCV_Ch-12_Obj I_15leku nzO;eku ds nks fi.M Ao B nks vyx&vyx nzO;eku jfgr fLizaxks a ftuds fLizax fu;rkad k1 o k2 gS ls yVdk;sx;s gSA ;fn nksuksa fi.M /okZ/kj bl izdkj nksyu djrs gS fd mues vf/kdre osx leku gS AoBds vk;keksa esa vuqikrgksxkA(A) k1 /k2 (B) 21 k/k (C) k2/k1 (D*) 12 k/k

16. A spring mass system oscillates with a frequency v. If it is taken in an elevator slowly accelerating upward,the frequency will(A) increase (B) decrease (C*) remain same (D) become zero,d fLizax nzO;eku fudk; vvko`fr ls nksyu dj jgk gSA ;fn bldks ij dh vksj /khes Roj.k ls xfr'khy fyV esays tk;k tk;s rks bldh vko`fr&(A) c



17. A spring-mass system oscillates in a car. If the car accelerates on a horizontal road, the frequency ofoscillation will(A) increase (B) decrease (C*) remain same (D) become zero.,d fLizax nzO;eku fudk; ,d dkj esa nksyu dj jgk gSA ;fn dkj {ksfrt lM+d ij Rofjr gksus yx tkrh gS] rksvkorZdky(A) c



2. A particle moves in a circular path with a uniform speed. Its motion is(A*) periodic (B) oscillatory(C) simple harmonic (D) angular simple harmonico`kkdkj iFk ij ,d d.k fu;r pky ls xfr'khy gSA bldh xfr gS&(A*)vkorhZ (B) nksyuh(C) ljy vkorhZ (D)dks.kh; ljy vkorhZ

3. A particle is fastened at the end of a string and is whirled in a vertical circle with the other end of the stringbeing fixed. The motion of the particle is(A*) periodic (B) oscillatory(C) simple harmonic (D) angular simple harmonic,d Mkjh ds fljs ij ,d d.k fpidk;k x;k gS rFkk Mkjh ds nwljs fljs dks fLFkj j[krs gq, bls m/okZ/kj o`k esa ?kqek;ktkrk gSA d.k dh xfr gSA(A*)vkorhZ (B) nksyuh(C) ljy vkorhZ (D)dks.kh; ljy vkorhZ

4. A particle moves in a circular path with a continuously increasing speed. Its motion is(A) periodic (B) oscillatory (C) simple harmonic (D*) none of themo`kkdkj iFk ,d d.k fujUrj c



(C) t = 0 ls r; dh xbZ nwjh ds(D) pky

11. A particle moves in the X-Y plane according to the equation

r = ( i

+ 2 j

) A cost.

The motion of the particle is(A*) on a straight line (B) on an ellipse (C*) periodic (D*) simple harmonicfuEu lehdj.k ds vuqlkj ,d d.k X-Y ry esa ljy vkorZ xfr dj jgk gSA

r = ( i

+ 2 j

) A cost.

d.k dh xfr gS&(A*) ljy js[kk ds vuqfn'k (B) nh/kZo`k ds vuqfn'k (C*)vkorhZ (D*) ljy vkorhZ

Sol. r

= ( i + 2 j ) A cos tThis is combination of S.H.M. along x and y axis having equationsx = A cos t & y = 2A cos tThis is equation of S.H.M. along a sraight line y = 2x.S.H.M. is always periodic.

r

= ( i + 2 j ) A cos t

;g x vkSj y v{k ds vuqfn'k ljy vkorZ xfr;ksa dk la;kstu gS ftuds lehdj.kx = A cos t vkSj y = 2A cos t;g ljy js[kk y = 2x ds vuqfn'k ljy vkorZ xfr dk lehdj.k gSAl0 vk0 x0 lnSo vkorhZ gksrk gSaA

12. A particle moves on the X-axis according to the equation x = x0 sin2 t. The motion is simple harnomic(A) with amplitude x0 (B) with amplitude 2x0

(C) with time period

2(D*)with time period

,d d.k lehdj.k x = x0 sin2 ds vuqlkj x-v{k ds vuqfn'k xfr'khy gSA xfr ljy vkorhZ gS&(A) x0vk;ke ds lkFk (B) 2x0vk;ke ds lkFk

(C) vkorZdky ds lkFk

2(D*vkorZdky ds lkFk

13. In a simple harmonic motion(A) the potential energy is always equal to the kinetic energy(B) the potential energy is never equal to the kinetic energy(C) the average potential energy in any time interval is equal to the average kinetic energy in that time interval(D*) the average potential energy in one time period is equal to the average kinetic energy in this period.ljy vkorZ xfr esa&(A) fLFkfrt tkZ lnSo xfrt tkZ ds cjkcj jgrh gS(B) fLFkfrt tkZ lnSo xfrt tkZ ls(C) fdlh le;kajky esa vkSlr fLFkfrt tkZ ml le;karjky esa vkSlr xfrt tkZ ds cjkcj jgrh gSA(D*) ,d vkorZkdy esa vkSlr fLFkfrt tkZ bl vkorZdky esa vkSlr xfrt tkZ ds cjkcj gksrh gSA

14. In a simple harmonic motion(A*) the maximum potential energy equals the maximum kinetic energy(B*) the minimum potential energy equals the minimum kinetic energy(C) the minimum potential energy equals the maximum kinetic energy(D) the maximum potential energy equals the minimum kinetic energyljy vkorZ xfr esa&(A*) vf/kdre fLFkfrt tkZ] vf/kdre xfrt tkZ ds cjkcj gksrh gSA(B*) U;wure fLFkfrt tkZ] U;wure xfrt tkZ ds cjkcj gksrh gSA(C) U;wure fLFkfrt tkZ] vf/kdre xfrt tkZ ds cjkcj gksrh gSA(D) vf/kdre fLFkfrt tkZ] U;wure xfrt tkZ ds cjkcj gksrh gSA

15. An object is released from rest. The time is takes to fall through a distance h and the speed of the object asit falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon



and the experiment is repeated(A*) the measured times are same (B*) the measured speeds are same(C) the actual times in the fall are equal (D) the actual speeds are equal,d oLrq dks fojkeoLFkk ls fxjk;k tkrk gSA h nwjh rd fxjus esa yxk le; rFkk blh nwjh rd fxjus ij pky dk ekiuyksyd ?kM+h }kjk fd;k tkrk gSA ;fn lEiw.kZ midj.k dks pkan ij ys tk;k tk;s rFkk bl iz;ksx dks nksgjk;k tk;s&(A*) ekis x;s le; leku gksxsa (B*) ekih x;h pkysa leku gksxhaA(C) fxjus esa yxs okLrfod le; leku gksxsaA (D) okLrfod pkysa ,d leku gksxhaA

16. Which of the following will change the time period as they are taken to moon ?(A*) A simple pendulum (B*) A physical pendulum(C) a torsional pendulum (D) a spring mass systempUnzek ij ys tkus ij fuEu ls fdlds vkorZdky ifjofrZr gks tk;sxas&(A*) ljy yksyd (B*) HkkSfrd yksyd(C) ejksM+h nksyd (D) fLizax nzO;eku fudk;

WORKED OUT EXAMPLES1. The equation of particle executing simple harmonic motion is x = (5 m) sin

3t)s( 1 . Write down the

amplitude, time period and maximum speed. Also find the velocity at t = 1 s.Sol. Comparing with equation x = A sin (t + ), we see that

the amplitude = 5 m,

and time period =

2= 1s

2

= 2s.

The maximum speed = A = 5 m s1 = 5 m/s.

The velocity at time t = dtdx

= AA cos (t + )

At t = 1 s,

v = (5 m) ( s1) cos

5 = 25

m/s.

2. A block of mass 5 kg executes simple harmonic motion under the restoring force of a spring. The amplitudeand the time period of the motion are 0.1 m and 3.14 s respectively. Find the maximum force exerted by thespring on the block.

Sol. The maximum force exerted on the block is kA when the block is at the extreme position.

The angular frequency =T2

= 2 s1

The spring constant = k = m2= (5 kg) (4 s2) = 20 N/m.

Maximum force = kA = (20 N/m) (0.1 m) = 2 N.

3. A particle executing simple harmonic motion has angular frequency 6.28 s1 and ampitude 10 cm. Find (a)the time period, (b) the maximum speed, (c) the maximum acceleration, (d) the speed when the displacementis 6 cm from the mean position, (e) the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

Sol. (a) Time period =

2= 28.6

2s = 1 s.

(b) Maximum speed = A = (0.1 m) (6.28 s1)= 0.628 m/s.

(c) Maximum acceleration = A2= (0.1 m) (6.28 s1)2= 4 m/s2.



(d) v = 22 xA = (6.28 s1) 22 )cm6()cm10( = 50.2 cm/s.

(e) At t = 0, the velocity is zero i.e., the particle is at an extreme. The equation for displacementmay be written as

x = A cost.The velocity is v = A sin t.

At t = 61

s, v = (0.1 m) (6.28 s1) sin

628.6

= ( 0.628 m/s) sin 3

= 54.4 cm/s.

4. A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to godirectly from its mean position to half the amplitude.

Sol. Let the equation of motion be x = A sin t.At t = 0, x = 0 and hence the particle is at its mean position. Its velocity is

v = A cos t = A which is positive. So it is going towards x = A/2.The particle will be at x = A/2, at a time t where

2A

= A sin t

or, sin t = 1/2or, t = /6Here minimum positive value of

t is chosen because we are interested in finding the time taken by the

particle to directly go from x = 0 to x = A/2.

Thus, t =

6 = )T/2(6

=12T

.

5. A block of mass m hangs from a vertical spring of spring constant k. If it is displaced from its equilibriumposition, find the time period of oscillations.

Sol. Suppose the length of the spring is stretched by a length . The tension in the spring is k and this is theforce by the spring on the block. The other force on the block is mg due to gravity. For equilibrium, mg = k or = mg/k. Take this position of the block as x = 0. If the block is further displaced by x, the resultant

force is k

x

kmg

mg = kx.

Thus, the resultant force is proportional to the displacement. The motion is simple harmonic with a time

period T = 2km

.

We see that in vertical oscillations, gravity has no effect on time period. The only effect it has is to shift theequilibrium position by a distance mg/k as if the natural length is increased (or decreased if the lower end ofthe spring is fixed) by mg/k.

6. A particle suspended from a vertical spring oscillates 10 times per second.At the highest point of oscillationthe spring becomes unstretched. (a) Find the maximum speed of the block. (b) Find the speed when thespring is stretched by 0.20 cm. Take g = 2 m/s2.

Sol.(a) The mean position of the particle during vertical oscillations is mg/k distance away from its position when thespring is unstretched. at the highest point, i.e., at an extreme position, the spring is unstretched.



Hence the amplitude is

A =k

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

The angular frequency is

=mk

= 2 v = (20) s1 ........(ii)

or,km

= 24001

s2.

Putting in (i), the amplitude is

A =

222

2 sms

4001

= 4001

m = 0.25 cm.

The maximum speed = A= (0.25 cm) (20 s1) = 5 cm/s.

(b) When the spring is stretched by 0.20 cm, the block is 0.25 cm 0.20 cm = 0.05 cm above the meanposition. The speed at this position will be

v = 22 xA

= (20 s1) 22 )cm05.0()cm25.0( 15.4 cm/s.

7. The pulley shown in figure has a moment of inertia about its axis and mass m. Find the time period ofvertical oscillation of its centre of mass. The spring has spring constant k and the string does not slip over thepulley.

Sol. Let us first find the equilibrium position. For rotational equilibrium of the pulley, the tensions in the two stringsshould be equal. Only then the torque on the pulley will be zero. Let this tension be T. The extension of thespring will be y = T/k, as the tension in the spring will be the same as the tension in the string. Fortranslational equilibrium of the pulley,

2T = mg or, 2 ky = mg or, y = k2mg

the spring is extended by a distance k2mg

when the pulley is in equilibrium.

Now suppose, the centre of the pulley goes down further by a distance x. The total increase in the length ofthe string plus the spring 2x (x on the left of the pulley and x on the right).As the string has a constant length,the extension of the spring is 2x. The energy of the system is

U =21

2 +21

mv2 mgx +21

k2

x2k2

mg



=21

mr2

v2 + k8gm 22

+ 2 kx2.

As the system is conservative, dtdU

= 0, giving,

0 =

mr2

v dtdv

+ 4 kxv

or, dtdv

=

mr

xk4

2

or, a = 2x where 2 =

mr

k4

2

Thus, the centre of mass of the pulley executes a simple harmonic motion with time period.

T = 2 )k4/(mr2

.

8. The friction coefficient between the two blocks shown in figure is and the horizontal plane is smooth. (a) Ifthe system is slightly displaced and released, find the time period. (b) Find the magnitude of the frictionalforce between the blocks when the displacement from the meanposition is x. (c) What can be the maximumamplitude if the upper block does not slip relative to the lower block ?fp=k esa n'kkZ;s x;s nks xqVdksa ds e/; ?k"kZ.k xq.kkad gS ,oa {kSfrt ry fpduk gSA (a) ;fn fudk; FkksM+k lk foLFkkfir djdsNksM+ fn;k tkrk gS rks vkorZdky Kkr djksA (b)tc ek/; fLFkfr ls foLFkkiu x gS rks xqVdksa ds e/; ?k"kZ.k cy dk ifjek.kKkr djksA (c) ;fn fupys xqVds ds lkis{k ijh xqVdk ugha fQlyrk gS rks vf/kdre vk;ke fdruk gks ldrk gS \

[M.Bank(07-08)_HCV__Ch.12_WOE_Q.8]

Sol. (a) For small amplitude, the two blocks oscillate together. The angular frequency is

=mM

k

and so the time period T = 2k

mM.

(b) The acceleration of the blocks at displacement x from the mean position is

a = 2x =

mMkx

The resultant force on the upper block is, therefore,

ma =

mMmkx

This force is provided by the friction of the lower block. Hence, the magnitude of the frictional force is

mM|x|mk

(c) Maximum force of friction required for simple harmonic motion of the upper block ismMAmk

at the

extreme positions. But the maximum frictional force can only be mg. Hence

mMAmk

= mg or, A =

kg)mM(

9. The left block in figure collides inelastically with the right block and sticks to it. Find the amplitude of theresulting simple harmonic motion.



Sol. Assuming the collision to last for a small interval only, we can apply the principle of conservation of momentum.

The common velocity after the collision is2v

. The kinetic energy =21

(2m)2

2v

=

41

mv2. This is also the

total energy of vibration as the spring is unstretched at this moment. If the amplitude is A, the total energy

can also be written as21

kA2. Thus,

21

kA2 =41

mv2, giving A = k2m

v..

10. Describe the motion of the mass m shown in figure. The walls and the block are elastic.

Sol. The block reaches the spring with a speed v. It now compresses the spring. The block is decelerated due to

the spring force, comes to rest when21

mv2 =21

kx2and return back. It is accelerated due to the spring force

till the spring acquires its natural length. The contact of the block with the spring is now broken. At thisinstant it has regained its speed v (towards left) as the spring is unstretched and no potential energy issotred. This process takes half the period of oscillation i.e., k/m . The block strikes the left wall after atime L/v and as the collision is elastic, it rebounds with the same speed v. After a time L/v, it again reachesthe spring and the process is repeated. The block thus undergoes periodic motion with time period k/m

+vL2

.

11. A block of mass m is suspended from the ceiling of a stationary elevator through a spring of spring constantk. Suddenly, the cable breaks and the elevator starts falling freely. Show that block now executes a simpleharmonic motion of amplitude mg/k in the elevator.

Sol. When the elevator is stationary, the spring is stretched to support the block. If the extension is x, the tensionis kx which should balance the weight of the block.

Thus, x = mg/k. As the cable breaks, the elevator starts falling with acceleration g. We shall work in heframe of reference of the elevator. Then we have to use a psuedo force mg upward on the block. This force willbalance the weight. Thus, the block is subjected to a net force kx by the spring when it is at a distance xfrom the position of unstretched spring. Hence, its motion in the elevator is simple harmonic with its meanposition corresponding to the unstretched spring. Initially, the spring is stretched by x = mg/k, where thevelocity of the block (with respect to the elevator) is zero. Thus, the amplitude of the resulting simple harmonicmotion is mg/k.

12. The spring shown in figure is kept in a stretched position with extension x0 when the system is released.Assuming the horizontal surface to be frictionless, find the frequency of oscillation.

Sol. Considering the two blocks plus the spring as a system there is no external resultant force on the system.Hence the centre of mass of the system will remain at rest. The mean positions of the two simple harmonicmotions occur when the spring becomes unstretched. If the mass m moves towards right through a distance



x and the mass M moves towards left through a distance X before the spring acquires natural length,x + X = x0. .............(i)

x and X will be the amplitudes of the two blocks m and M respectively. As the centre of mass should notchange during the motion, we should also have

mx = MX. .............(ii)

From (i) and (ii), x = mMMx0

and X = mMmx0

.

Hence, the left block is x = mMMx0

distance away from its mean position in the beginning of the motion. The

force by the spring on thick block at this isntant is equal to the tension of spring i.e., T = kx0.

Now x =mM

Mx0

or,, x0 = MmM

x

Thus, T =M

)mM(k x or, a =

mT

=Mm

)mM(k x.

The angular frequency is, therefore, =Mm

)mM(k and the frequency is v =

2 = 21

Mm)mM(k

.

13. Assume that a narrow tunnel is dug between two diametrically opposite points of the earth. Treat the earth asa solid sphere of uniform density. Show that if a particle is released in this tunnel, it will execute a simpleharmonic motion. Calculate the time period of this motion.

Sol.

Consider the situation shown in figure. Suppose at an instant t the particle in the tunnel is at a distance xfrom the centre of the earth. Let us draw a sphere of radius x with its centre at the centre of the earth. Onlythe part of the earth within this sphere will exert a net attraction on the particle. Mass of this part is

M =3

3

R34

x34

M = 33

Rx

M.

The force of attraction is, therefore,

F = 233

xMm)R/x(G

= 3RGMm

x.

This force acts towards the centre of the earth. Thus, the resultant force on the particle is opposite to thedisplacement from the centre of the earth and is proportional to it. The particle, therefore, executes a simpleharmonic motion in the tunnel with the centre of the earth as the mean position.

The force constant is k = 3RGMm

, so that the time period is

T = 2km

= 2GMR3

14. A simple pendulum of length 40 cm oscillates with an angular amplitude of 0.04 rad. Find (a) the time period,(b) the linear amplitude of the bob, (c) the speed of the bob when the string makes 0.02 rad with the verticaland (d) the angular acceleration when the bob is in momentary rest. Take g = 10 m/s2.

Sol.(a) The angular frequency is

= /g =m4.0s/m10 2

= 5 s1



the time period is

2= 1s5

2

= 1.26 s.

(b) Linear amplitude = 40 cm 0.04 = 1.6 cm(c) Angular speed at displacement 0.02 rad is

= (5 s1) 22 )02.0()04.0( rad = 0.17 rad/s.

where speed of the bob at this constant= (40 cm) 0.1751 = 6.8 cm/s.

(d) At momentary rest, the bob is in extreme position.Thus, the angular acceleration

= (0.04 rad) (25 s2) = 1 rad/s2.

15. A simple pendulum having a bob of mass m undergoes small oscillations with amplitude 0. Find the tensionin the string as a function of the angle made by the string with the vertical. When is this tension maximum,and when is it minimum ?

Sol. Suppose the speed of the bob at angle is v. Using conservation of energy between the extreme position andthe position with angle ,

21

mv2 = mg (cos cos0). .............(i)

As the bob moves in a circle path, the force towardsthe centre should be equal to mv2/. Thus,

T = mg cos = mv2/.Using (i),

T mg cos = 2mg (cos cos0)or, T = 3 mg cos 2 mg cos0.Now cos is maximum at = 0 and decreases as || increases (for || < 90).Thus, the tension is maximum when = 0 i.e., at the mean position and is minimum when = 0 i.e., at extreme positions.

16. A simple pendulum is taken at a place where it separation from the earths surface is equal to the radius ofthe earth. Calculate the time period of small oscillations if the length of the string is 1.0m. Take g = 2 m/s2at the surface of the earth.

Sol. At a height R (radius of the earth) the acceleration due to gravity is

g = 2)RR(GM

=41

2RGM

= g/4.

The time period of small oscillations of the simple pendulum is

T = 2 g/ = 2 22 s/m41

m0.1

= 2

s2 = 4s.

17. A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If theacceleration is a0 and the length of the pendulum is , find the time period of small oscillations about themean position.

Sol. We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a psuedoforce ma0 on the bob of mass m.



For mean position, the acceleration of the bob with respect to the car should be zero. If be the angle madeby the string with the vertical, the tension, weight and the psuedo force will add to zero in this position.

Suppose, at some instant during oscillation, the string is further deflected by an angle so that t h edisplacement of the bob is x. Taking the components perpendicular to the string,component of T = 0,component of mg = mg sin ( + ) andcomponent of ma0 = ma0 cos ( + ).Thus, the resultant component F

= m [g sin ( + ) a0 cos ( + )]Expanding the sine and cosine and putting cos 1,sin = x/, we get

F = m

x)sinacosg(cosasing 00 . ...........(i)

At x = 0, the force F on the bob should be zero, as this is the mean position. Thus by (i),0 = m[g sin a0 cos ] ..........(ii)

giving tan = ga0

Thus, sin =22

0

0

ga

a

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

cos = 220 ga

g

...........(iv)

Putting (ii), (iii) and (iv) in (i), F = m 202 ag

x

or, F = m 2 x, where 2 =

20

2 ag

This is an equation of simple harmonic motion with time period

t =

2= 2 4/12

02 )ag(

An easy working rule may be found out as follows. In the mean position, the tension, the weight and thepsuedo force balance.From figure, the tension is

T = 220 )mg()ma(

or,mT

= 220 ga .

This plays the role of effective g. Thus the time period is

t = 2m/T = 2 4/12

02 ]ag[

.

18. A uniform meter stick is suspended through a small pin hole at the 10 cm mark. Find the time peiod of smalloscillation about the point of suspension.

Sol. Let the mass of the stick be m. The moment of inertia of the stick about the axis of rotation through the pointof suspension is



=12

m 2 + md2,

where = 1 m and d = 40 cm.

The separation between the centre of mass of the stick and the point of suspension is d = 40 cm. The timeperiod of this physical pendulum is

T = 2 mgd

= 2 )mgd/(md12m 22

= 2

4/16.0

121

s = 1.55 s.

19. The moment of inertia of the disc used in a torsional pendulum about the suspension wire is 0.2 km-m2. Ifoscillates with a period of 2s. Another disc is placed over the first one and the time period of the systembecomes 2.5 s. Find the moment of inertia of the second disc about the wire.

Sol. Let the torsional constant of the wire be k. The moment of inertia of the first disc about the wire is 0.2 kg-m2.Hence the time period is

2s = 2k

= 2k

mkg2.0 2 ........... (i)

When the second disc having moment of inertia 1 about the wire is added, the time period is

2.5 s = 2kmkg2.0 1

2 ..............(ii)

From (i) and (ii),425.6

= 21

2

mkg2.0mkg2.0

This gives 1 0.11 kg m2.

20. A uniform rod of mass m and length is suspended through a light wire of length and torsional constant kas shown in figure. Find the time period of the system makes (a) small oscillations in the vertical plane aboutthe suspension point and (b) angular oscillations in the horizontal plane about the centre of the rod.

Sol.(a) The oscillations take place about horizontal line through the point of suspension and perpendicular to theplane of the figure. The moment of inertia of the rod about this line is

12m 2 + m2 =

1213

m2.

The time period = 2 mg

= 2

mg12m13 2

= 2 g1213

.

(b) The angular oscillations take place about the suspension wire. The moment of inertia about this line is m2/12. The time period is



2k

= 2 k12m 2

.

21. A particle is subjected to two simple harmonic motionsx1 = A1 sin t

and x2 = A2 sin (t + /3).Find (a) the displacement at t = 0, (b) the maximum speed of the particle and (c) the maximum accelerationof the particle.

Sol.(a) At t = 0, x1 = A1 sin t = 0and x2 = A2 sin (t + /3)

= A2 sin (/3) = 23A2 .

Thus, the resultant displacement at t = 0 is

x = x1 + x2 = A2 23

(b) The resultant of the two motions is a simple harmonic motion of the same angular frequency. The amplitudeof the resultant motion is

A = )3/cos(AA2AA 2122

21

= 2122

21 AAAA .

The maximum speed is

umax = A = 2122

21 AAAA

(c) The maximum acceleration is

amax = A 2 = 2 21

22

21 AAAA .

22. A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes andequal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, find the phasedifference between the individual motions. HCV_Ch-12_WOE _22,d d.k nks ljy vkorZ xfr;ksa ftuds vk;ke o vko`fk leku gS ds vUrxZr xfreku gSA ;fn ifj.kkeh vk;kevyx&vyx xfr;ksa ds vk;ke ds cjkcj gS rks vyx&vyx xfr;ksa ds chp dykUrj Kkr djksA

Sol. Let the amplitudes of the individual motions be A each. The resultant amplitude is also A. If the phasedifference between the two motions is ,

A = cos.A.A2AA 22

or, = A )cos1(2 = 2A cos2

or, cos2

=21

or, = 2/3.

EXERCISE1. A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At

t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement xat time t. Find the magnitude of the acceleration of the particle at t = 4 s.,d d.k 10 lseh vk;ke rFkk 6 lsd.M vkorZdky ds lkFk ljy vkorZ xfr dj jgk gSA /kukRed x fn'kk esa tkrsgq,] t = 0 ij bldh fLFkfr x = 5 lseh gSA le; t ij foLFkkiu x-ds fy;s lehdj.k fyf[k;sA t = 4 lsd.M ijd.k ds Roj.k dk ifjeki Kkr dhft;sA



1. x = (10 cm) sin

6t

s62

, 11 cm/s2

2. The position, velocity and acceleration of a particle executing simple hamonic motion are found to havemagnitudes 2 cm, 1 m/s and 10 m/s2 at a certain instant Find the amplitude and the time period of themotion.ljy vkorZ xfr dj jgs d.k dh fdlh fof'k"V {k.k ij fLFkfr osx rFkk Roj.k ds ifjek.k e'k% 2lseh, 1 eh- /ls- 10eh-/ls-2 gSa xfr dk vk;ke rFkk vkorZdky Kkr dhft;sA2. 4.9 cm, 0.28 s

3. The particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the meanposition are the kinetic and potential energies equal ?,d d.k 10 lseh vk;ke ds lkFk ljy vkorZxfr dj jgk gSA ek/; fLFkfr ls fdl nwjh ij xfrt rFkk fLFkfrt tkZ,cjkcj gksxh A3. 5 2

4. The maximum speed and acceleration of a particle executing simple harmonic motion are10 cm/s and 50 cm/s2. Find the position (s) of the particle when the speed is 8 cm/s.ljy vkorZxfr dj jgs d.k dh vf/kdre pky rFkk Roj.k 10 lseh/ls- ,oa 50 lseh/ls2.gSA tc d.k dh pky 8 lseh@ls- gks rks d.k dh fLFkfr fLFkfr;k Kkr dhft;sA4. 1.2 cm from the mean position

5. A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s1) t+ /6]. Find (a)the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration att = 0.10 xzke nzO;eku dk d.k lehdj.k x = (2.0 cm) sin [(100 s1) t+ /6]ds vuqlkj nksyu dj jgk gSA Kkr dhft;s%(a) vk;ke] vkorZdky rFkk fLizax fu;rkad (b) t = 0 ij fLFkfr] osx rFkk Roj.k5. (a) 2.0 cm, 0.063 s, 100 N/m (b) 1.0 cm, 1.73 m/s, 100 m/s2

6. The equation of motion of a particle started at t = 0 is given by x = 5 sin (20 t + /3) where x is in centimetreand t in second. When does the particle(a) first come to rest(b) first have zero acceleration(c) first have maximum speed ?t = 0 ij xfr izkjEHk djus okys ,d d.k dh xfr dh lehdj.k x = 5 sin (20 t + /3) }kjk O;Dr dh tkrh gS] tgkx lseh esa rFkk t lsd.M esa gSA d.k(a) igyh ckj dc fojkekoLFkk esa vk;sxk(b) dk Roj.k igyh ckj dc 'kwU; gksxk(c) dh pky igyh ckj dc vf/kdre gksxh \

6. (a) 120

s (b) 30

s (c) 30

s

7. Consider a particle moving in simple harmonic motion according to the equationx = 2.0 cos (50 t + tan1 0.75)where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come torest for the first time ? (b) When does the acceleration have its maximum magnitude for the first time ? (c)When does the particle come to rest for the second time ?,d d.k fuEufyf[kr lehdj.k ds vuqlkj ljy vkorZ xfr dj jgk gSAx = 2.0 cos (50 t + tan1 0.75);gk x lseh esa rFkk t lsd.M esa gSA xfr t = 0 ij izkjEHk gksrh gSA (a)d.k igyh ckj fojkekoLFkk esa dc vk;sxk\ (b)igyh ckj Roj.k dk ifjek.k vf/kdre dc gksxk\ (c)d.k nwljh ckj fojkekoLFkk esa dc vk;sxk\7. (a) 1.6 102 s (b) 1.6 102 s (c) 3.6 102 s

8. Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to changevalue from half the amplitude to the amplitude.TvkorZdky ls gks jgh ljy vkorZ xfr ij fopkj dhft;sA foLFkkiu dk eku vk;ke ds vk/ks ls vk;ke rd ifjofrZr



gksus esa yxs le; dh x.kuk dhft;sA8. T/6

9. The pendulum of a clock is replaced by a spring - mass system with the spring having spring constant 0.1 N/m. What mass should be attached to the spring ?,d ?kM+h dk yksyd] fLizax&nzO;eku fudk; ls izfrLFkkfir dj fn;k x;k gS] fLizax dk fu;rkad 0.1 U;wVu@eh- gSA fLizaxls fdruk nzO;eku tksM+uk pkfg;s\9. 10 g

10. A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals thelength of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block.,d m/okZ/kj fLizax ls yVdk;k x;k xqVdk lkE;koLFkk esa gSA O;Dr dhft;s fd fLizax dk izlj.k] ,d rqY; ljy yksyddh yEckbZ ds cjkcj gS vFkkZr~ CykWd dh vko`fk ds cjkcj vko`fk okys yksyd dh yEckbZA

11. A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 mand time period 0.314 s. Find the maximum force exerted by the spring on the block.11. 25 N,d m/okZ/kj fLizax ls yVdk;k x;k 0.5 fdxzk nzO;eku dk xqVdk ljy vkorZ xfr dj jgk gSA ftldk vk;ke0.1 eh- rFkk vkorZdky 0.314 lsd.M gSA fLizax }kjk xqVds ij yxk;k x;k vf/kdre cy Kkr dhft;sA

12. A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4s. Ifthe oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.,d m/okZ/kj fLizax ls yVdk;h x;h 2 fdxzk nzO;eku okyh oLrq] ljy vkorZ xfr dj jgh gS] bldk vkorZdky 4lsd.MgSA ;fn nksyu jksd fn;s tk;s rFkk oLrq lkE;koLFkk esa yVdh jgs] fLizax esa lafpr fLFkfrt tkZ Kkr dhft;sA12. 40 J

13. A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A blockfastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations eachsecond, what is the mass of the block ?,d fLizax dks 25 lseh [khapus ij blesa 5 twy tkZ lafpr gksrh gSA bldks m/okZ/kj j[kk x;k gS rFkk bldk fupykfljk fLFkj gSA nwljs fljs ij ,d xqVdk fpidk;k x;k gS tks vYi vk;ke okys nksyu dj jgk gSA ;fn xqVdk izR;sdlsd.M esa 5 nksyu dj jgk gS] xqVds dk nzO;eku fdruk gS\13. 0.16 kg

14. A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of springconstant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smallerblock when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on thesmaller block at this position. When is this force smallest in magnitude? (c) What can be the maximumamplitude with which the two blocks may oscillate together?m nzO;eku okyk ,d NksVk xqVdk]M nzO;eku okys cM+s xqVds ij j[kk gqvk gS] tks fp=kkuqlkj ,d k cy fu;rkad okyhm/okZ/kj fLizax ls tqM+k gqvk gSA fudk; m/okZ/kj nksyu dj jgk gSA (a)tc bldks lkE;koLFkk ls ij dh vksj x nwjhrd foLFkkfir fd;k tkrk gS] NksVs xqVds ij ifj.kkeh cy Kkr dhft;sA (b) bl fLFkfr esa NksVs xqVds ij vfHkyEcor~cy Kkr dhft;sA ;g cy U;wure ifjek.k dk dc gksxk\ (c)vk;ke dk vf/kdre eku fdruk gks ldrk gS fd nksuksaxqVds ,d lkFk nksyu djrs jgsA

14. (a)mM

mkx

(b) mg mM

mkx

, at the highest point (c) gk

)mM(

15. The block of mass m1 shown in figure is fastened to the spring and the block of mass m2 is placed against it.(a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a furtherdistance (2/k) (m1 + m2) g sin against the spring and released. Find the position where the two blocksseparate. (c) What is the common speed of blocks at the time of separation?fp=k esa iznf'kZr fd;k x;k m1 nzO;eku dk CykWd fLizax ls fpidk;k x;k gS rFkk m2 nzO;eku okyk CykWd blls fVdkdj j[kk x;k gSA (a) lkE;koLFkk esa fLizax dk lEihM+u Kkr dhft;sA (b) CykWdksa dks (2/k) (m1 + m2) g sin nwjh rdvksj foLFkkfir djds NksM+ fn;k tkrk gSA og fLFkfr Kkr dhft;s] tgk nksuksa CykWd vyx gks tk;saxsA (c)vyx gksrsle; nksuks a CykWdksa dh mHk;fu"B pky fdruh gksxh\



15. (a)k

sing)mm( 21 (b) when the spring acquires its natural length (c) sing)mm(k3

21

16. In figure k = 100 N/m, M = 1 kg and F = 10 N. (a) Find the compression of the spring in the equilibriumposition. (b)A sharp blow by some external agent imparts a speed of 2 m/s to the block towards left. Find thesum of the potential energy of the spring and the kinetic energy of the block at this instant. (c) Find the timeperiod of the resulting simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of thespring when the block is at the left extreme. (f) Write the potential energy of the spring when the block is atthe right extreme.The answers of (b), (e) and (f) are different. Explain why this does not violate the principle of conservation ofenergy.fp=k esa k = 100 U;wVu/eh-, M = 1 fdxzk rFkk F = 10 U;wVu gSA (a) lkE;koLFkk esa fLizax dk lEihM+u Kkr dhft;sA (b)fdlh ck ;qfDr }kjk ,dne yxk;s x;s /kDds ds dkj.k CykWd cka;h vksj 2 eh-/ls- dh pky izkIr dj ysrk gSA bl{k.k ij fLizax dh fLFkfrt tkZ rFkk CykWd dh xfrt tkZ dk ;ksx dhft;sA (c) ifj.kkeh ljy vkorhZ xfr dkvkorZdky Kkr dhft;sA (d)vk;ke Kkr dhft;s (e)tc CykWd cka;h vksj vafre fLFkfr esa gks rks fLizax dh fLFkfrt tkZfyf[k;sA (f) CykWd tc nk;ha vksj vafre fLFkfr esa gks fLizax dh fLFkfrt tkZ fyf[k;sAThe answers of (b), (e) and (f) are different. Explain why this does not violate the principle of conservation ofenergy.

16. (a) 10 cm (b) 2.5 J (c) /5 s (b) 20 cm (e) 4.5 J (f) 0.5 J

17. Find the time period of the oscillation of mass m in figures a, b, c. What is the equivalent spring constant ofthe pair of springs in each case ?fp=k (a, b, c) esa iznf'kZr m nzO;eku ds nksyu dk vkorZdky Kkr dhft;sA izR;sd fLFkfr esa fLizaxksa ds tksM+s dk rqY;fLizax fu;rkad fdruk gksxk\

(a) (b) (c)

17. (a) 221 kk

m (b) 2 21 kk

m (c) 2 21

21kk

)kk(m

18. The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M.If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block (b)the energy stored in the spring when the block passes through the equilibrium position and (c) the kineticenergy of the block at this position.tc ,d O;fDr Mksjh [khapuk izkjEHk djrk gS rks fp=k esa iznf'kZr fLizax fcuk [khaph gqbZ fLFkfr esa gSA CykWd dk nzO;ekuM gSA ;fn O;fr }kjk yxk;k x;k fu;r cy F gS] Kkr dhft;s A (a) CykWd dh xfr dk vk;ke rFkk vkorZdky (b)tcCykWd lkE;koLFkk ls xqtjrk gS rks fLizax esa lafpr tkZ rFkk (c) bl fLFkfr esa CykWd dh xfrt tkZ

18. (a)kF

, 2kM

, (b)k2

F2 (c)k2

F2

19. A particle of mass m is attatched to three springs A, B and C of equal force constant k as shown in figure. Ifthe particle is pushed slightly against the spring C and released, find the time period of oscillations.tSlkfd fp=k esa iznf'kZr fd;k x;k gSA leku cy fu;rkad k okyh rhu fLizaxks aA, B rFkk C lsm nzO;eku dk ,d d.k



tqM+k gqvk gSaA ;fn d.k dks fLizax Cds fo:) FkksMk lk foLFkkfir djds NksM+ fn;k tkrk gS] nksyu dk vkorZdky Kkrdhft;sA

19. 2k2

m

20. Repeat the previous exercise if the angle between each pair of springs is 120 initially.;fn izR;sd fLizax ;qXe ds chp izkjfEHkd dks.k 120 gks rks fiNys iz'u% dks nksgjkb;sA

20. 2 k3m2

21. The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. Theman exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.tc dksbZ O;fr CykWd dks [khpuk 'kq: djrk gS rks fp=k esa iznf'kZr leLr fLizaxsa] izkjEHk esa fcuk [khph gbZ fLFkfr esagSA O;fDr CykWd ij ,d fu;r cy Fyxkrk gSA CykWd dh xfr dk vk;ke rFkk vkof`r Kkr dhft;sA

21.133221

32kkkkkk

)kk(F

,

21

)kk(Mkkkkkk

32

133221

22. Find the elastic potential energy stored in each spring shown in figure, when the block is in equilibrium. Alsofind the time period of vertical oscillation of the block.fp=k esa iznf'kZr CykWd tc lkE;koLFkk esa gS] izR;sd fLizax esa lafpr izR;kLFk fLFkfrt tkZ Kkr dhft;sA

22.1

22

k2gM

,2

22

k2gM

and3

22

k2gM

from above, time period = 2

321 k1

k1

k1M

23. The string, the spring and the pulley shown in figure are light. Find the time period of the mass m.fp=k esa Mksjh] fLizax rFkk f/kjuh ux.; nzO;eku dh gSA m nzO;eku dk vkorZdky Kkr dhft;sA

23. 2km



24. Solve the previous problem if the pulley has a moment of inertia about its axis and the string does not slipover it.fiNys iz'u dks gy dhft;s] ;fn f?kjuh dk bldh v{k ds ifjr% tMRo vk?kw.kZ gS rFkk Mksjh bl ij fQlyrh ughgSA

24. 2k

)r/m( 2

25. Consider the situation shown in figure. Show that if the blocks are displaced slightly in opposite directionsand released, they will execute simple harmonic motion. Calculate the time period.fp=k esa iznf'kZr O;oLFkk ij fopkj dhft;sA ;fn CykWdksa dks foijhr fn'kkvksa esa FkksM+k lk foLFkkfir djds NksM+ fn;k rksO;Dr dhft;s fd os ljy vkorZ xfr djsxasA vkorZdky dh x.kuk dhft;s A

25. 2 k2m

26. A rectangular plate of sides a and b is suspended from a ceiling by two parallel strings of length L each(figure). The separation between the strings is d. The plate is displaced slightly in its plane keeping thestrings tight. Show that it will excute simple harmonic motion. Find the time period.Nr ls yVrh gqbZ nks lekukarkj Mksfj;ksa ls ,d vk;rkdkj IysV yVdkbZ x;h gS] bldh Hkqtk, a rFkk b gSA izR;sd Mksjhdh yEckbZ L gSA Mksfj;ksa ds chp dh nqjh d gSA Mksfj;kas dks ruk gqvk j[krs gq, IysV dks blds ry esa FkksM+k lk foLFkkfirfd;k tkrk gSA O;Dr dhft;s fd ;g ljy vkorZ xfr djsxhA vkorZdky Kkr dhft;sA

26. 2p gL

27. A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface underthe restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at theinstant it passes through the mean position. Assuming that the two blocks move together, find the frequencyand the amplitude of the motion.100 U;wVu@ eh- cy fu;rkad okyh fLiazx ds izR;ku;u cy ds izHkko esa ,d fpduh {ksfrt lrg ij j[kk gqvk 1 fdxzknzO;eku dk CykWd ljy vkorZ xfr dj jgk gSA tc ;g ek/; fLFkfr ls xqtjrk gS] blds ij ,d 3 fdxzk nzO;ekudk CykWd /khjs ls j[k fn;k tkrk gSA ;g ekurs gq, fd nksuks CykWd ,d lkFk xfr djrs gS] xfr dh vkof`r ,oa vk;keKkr dhft;s A

27.2

5Hz, 5 cm

28. The left block in figure moves at a speed v towards the right block placed in equilibrium. All collisions to takeplace are elastic and the surface are frictionless. Show that the motions of the two blocks are periodic. Findthe time period of these periodic motions. Neglect the widths of the blocks.fp=k esa iznf'kZr ck;ka CykWd] lkE;koLFkk esa fLFkr nk;sa CykWd dh vksj vpky ls xfr'khy gSA gksus okyh leLr VDdjsizR;kLFk gS rFkk lrgsa ?k"kZ.k jfgr gSA O;Dr dhft;s fd nksuksa Cykdks dh xfr vkorhZ gksxhA budh vkorhZ xfr;ksa dkvkorZdky Kkr dhft;s] CykWdks dh pkSMkbZ;k ux.; eku yhft;s A



28.

vL2

km

29. Find the time period of the motion of the particle shown in figure. Neglect the small effect of the bend near thebottom.fp=k esa iznf'kZr d.k dh xfr dk vkorZdky Kkr dhft;sA iSnsa ds fudV eksM+ dk izHkko ux.; eku yhft;sA

29. 0.73 s

30. All the surfaces shown in figure are frictionless. The mass of the car is M, that of the block is m and thespring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through alength x0 when the system is released. (a) Find the amplitudes of the simple harmonic motion of the blockand of the car as seen from the road. (b) Find the time period (s) of the two simple harmonic motions.fp=k esa iznf'kZr leLr lrgs ?k"kZ.k jfgr gSA dkj dk nzO;ekuM, CykWd dk nzO;ekum rFkk fLizax dk fLizax fu;rkadk gS A izkjEHk esa tc fudk; dks eqDr fd;k x;k dkj rFkk CykWd fojkekoLFkk esa gS ,oa fLizax x0yEckbZ ls [khph gqbZgSA (a)lM+d ls ns[kus ij CykWd rFkk dkj ds vk;keksa dk eku Kkr dhft;s (b) nksuksa ljy vkorZ xfr;ksa dk vkorZdky(ds vkorZdky) Kkr dhft;sA

30. (a)mM

Mx0

,mM

mx0

(b) 2 )mM(kmM

31. A uniform plate of mass M stays horizontally and symmetricaly on two wheels rotating in opposite directions(figure). The separation between the wheels is L. The friction coefficient between each wheel and the plate is. Find the time period of oscillation of the plate if it is slightly displaced along its length and released.foifjr fn'kkvksa esa ?kwf.kZr nks ifg;ksa ijM nzO;eku dh ,d le:i IysV {ksfrt :i ls j[kh gqbZ gSA ifg;s ds e/; varjkyLgSA izR;sd ifg;s rFkk IysV ds e/; ?k"kZ.k xq.kkad gS A ;fn IysV dks bldh yEckbZ ds vuqfn'k FkksM+k lk foLFkkfirdjds NksM+ fn;k tk;s rks blds nksyu dk vkorZdky Kkr dhft;sA

31. 2 g2

32. A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulumclocks are of this type. Find the length of a seconds pendulum at a place whereg = 2 m/s2.nks lsd.M vkorZdky okyk yksykd] ^^ lsd.M+ yksyd ^^ dgykrk gSA ?kfM+;ksa esa bl izdkj ds yksyd iz;qDr fd;s tkrsgSA ;fn fdlh LFkku ij g = 2 m/s2 gS] ogkW ij lsd.M yksyd dh yEckbZ Kkr dhft;s A32. 1 m

33. The angle made by the string of a simple pendulum with the vertical depends on time as = 90

sin [ s1)

t]. Find the length of the pendulum if g = 2 m/s2.

ljy yksyd dh Mksjh dk m/okZ/kj ls dks.k le; ij = 90

sin [ s1) t]ds vuqlkj fuHkZj djrk gSA ;fn g



= 2 m/s2 gS rks yksyd dh yEckbZ Kkr dhft;s\33. 1m

34. The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours ?fdlh fof'k"V ?kM+h ds yksyd dk vkorZdky 2.04ls- gSA 24 ?kaVs i'pkr~ ;g fdlh rst vFkok /kheh pysxh ?

34. 28.8 minutes slow

35. A pendulum clock giving correct time at a place where g = 9.8 m/s2 is taken to another place where it loses24 seconds during 24 hours. Find the value of g at this new place.fdlh LFkku ij tgk g = 9.8 m/s2 gSA yksyd ?kMZ lgh le; n'kkZrh gSA bldks fdlh vU; LFkku ij ys tkus ij ;g24 ?kaVks esa 24 lsd.M ihNs jg tkrh gSA bl u;s LFkku ij g dk eku Kkr dhft;sA35. 9.795 m/s2

36. A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes smalloscillations. (a) How many oscillations does it make per second ? (b) What will be the frequency if thesystem is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s2.,d 5.0 m yEch Mksjh ls ,d Hkkjh xasn yVdkdj ljy yksyd cuk;k x;k gSA ;g vYi vk;ke ds nksyu dj jgk gSA(a) ;g izfr lsd.M fdrus nksyu djsxk ? (b) ;fn bl fudk; dks pUnzek ij ys tk;k tk;s tgk xq:Roh; Roj.k dkeku 1.67 m/s2 gS] bldh vko`fr fdruh gksxh \36. (a) 0.70/ (b) 1/(2/ 3 ) Hz

37. The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find theangular amplitude.nksyu dj jgs ,d yksyd dh Mksjh esa vf/kdre ruko] blds U;wure ruko dk nqxuk gSA dks.kh; vk;ke Kkr dhft;sA37. cos1 (3/4)

38. A small block oscillates back and forth on a smooth concave surface of radius R (figure). Find the time periodof small oscillation.R f=kT;k dh ,d fpduh vory lrg ij ,d NksVk CykWd vkxs&ihNs nksyu dj jgk gSA (fp=k)vYi vk;ke ds nksyuksadk vkorZdky Kkr fdft;sA

38. 2 g/R

39. A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. Itmakes small oscillations about the lowest point. Find the time period.cgqr vf/kd f=kT;k Rdh ,d [kqjnjh vory lrg ij m nzO;eku dh xksykdkj xasn fcuk fQlys yq



ls iz{ksfir fd;k tk;s A

41.2

gR

in each case

42. Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earthscentre where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational forceexerted by the earth on a particle of mass m placed in the tunnel at a distance x from the centre of the tunnel.(b) Find the component of this force along the tunnel and perpendicular to the tunnel. (c) Find the normalforce exerted by the wall on the particle. (d) find the resultant force on the particle. (e) Show that the motionof the particle in the tunnel is simple harmonic and find the time period.eku yhft;s fd i`Foh ds dsUnz lsR/2 yEcor~ nwjh ij fLFkr] thok ds vuqfn'k ,d lqjax [kksnh tkrh gS] ;gk R i`Fohdh f=kT;k gSA (a) lqjax ds dsUnz ls x nwjh ij fLFkr m nzO;eku ds d.k ij i`Foh ds dkj.k yxus okys xq:Roh; cydk eku Kkr dhft;sA (b) lqjax ds vuqfn'k ,oa lqjax ds yEcor~ bl cy ds ?kVd Kkr dhft;sA (c) nhokj }kjk d.kij yxk;k x;k vfHkyEcor~ cy Kkr fdft;sA (d)nhokj ij ifj.kkeh cy Kkr dhft;sA (e) O;Dr dhft;s fd lqjax esad.k dh xfr ljy vkorhZ gS] bldka vkorZdky Hkh Kkr dhft;sA

42. (a) 3RGMm

4/Rx 42 (b) 3RGMm

x, 2R2GMm

(c) 2R2GMm

, (d) 3RGMm

x (e) 2 )GM/(R3

43. A simple pendulum of length is suspended through the ceiling of an elevator. Find the time period of smalloscillations if the elevator (a) is going up with an acceleration a0 (b) is going down with an acceleration a0 and(c) is moving with a uniform velocity.,d fyV dh Nr ls yEckbZ dk ljy yksyd yVdk;k x;k gSA vYi vk;ke ds nksyuksa dk vkorZdky Kkr dhft;s];fn fyV (a)ij dh vksj a0 Roj.k ls xfr'khy gSaA (b) uhps dh vksj a0 Roj.k ls xfr'khy gS] rFkk (c) ,d leku osxls xfr'khy gSA

43. (a) 20ag

(b) 2

0ag

(c) 2 g

44. A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes /3 seconds to completeone oscillation. Find the acceleration of the elevator.,d fyV dh Nr ls yVdk;k x;k 1QqV yEckbZ dk ljy yksyd /3 lsd.M esa ,d nksyu iwjk dj ysrk gSA fyVdk Roj.k Kkr dhft;sA

44. 4 feet/s2 upwards45. A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a

horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making anapproximate analysis, find the acceleration of the car.,d dkj dh Nr ls yVdk;s x;s ljy yksyd dk vkorZdky 4 lsd.M gS] tcfd dkj {ksfrt lMd ij ,d leku :ils xfr'khy gSA tc ,DlhysVj nck;k tkrk gSA bldk vkorZdky ifjofrZr gksdj 3.99 lsd.M gks tkrk gSA vuqekfurfo'ys"k.k djds] dkj dk Roj.k Kkr dhft;s A45. g/10

46. A simple pendulum of length is suspended from the ceiling of a car moving with a speed v on a circularhorizontal road of radius r. (a) Find the tension in the string when it is at rest with respect to the car. (b) Findthe time period of small oscillation.r f=kT;k dh ,d {ksfrt o`kkdkj lM+d ij ,d dk v pky ls xfr'khy gS] bldh Nr ls yEckbZ dk ,d ljy yksydyVdk;k x;k gSA (a) tc ;g dkj ds lkis{k fLFkj gS] Mksjh esa ruko Kkr dhft;sA (b)vYi vk;ke okys nksyuksa dkvkorZdky Kkr dhft;s A

46. (a) ma (b) 2 a/ where a =2/1

22

rvg

47. The ear-ring of a lady shown in figure. has a 3 cm long light suspension wire. (a) Find the time period of smalloscillations if the lady is standing on the ground. (b) The lady now sits in a merry-go-round moving at 4 m/sin a circle of radius 2 m. Find the time period of small oscillations of the ear-ring.fp=k esa iznf'kZr L=kh ds dku dk >qeds esa 3 lseh yEck ,d gYdk rkj gSA (a) ;fn L=kh tehu ij [kM+h gqbZ gS rks vYivk;ke ds nksyuksa dk vkorZdky Kkr fdft;s A (b) ;fn ;g L=kh tc ^^ esjh&xks&jkm.M^^ esa cSB tkrh gS tks 4 eh-@ls pky ls py jgk gSA >qeds ds vYi vk;ke ds nksyuksa dk vkorZdky Kkr dhft;sA



47. (a) 0.34s (b) .30 s

48. Find the time period of small oscillations of the following systems. (a) A metre stick suspended through the20 cm mark. (b) A ring of mass m and radius r suspended through a point on its perphery. (c) A uniformsquare plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspendedthrough a point r/2 away from the centre.fuEu fudk;ksa ds fy;s vYi vk;ke ds nksyuksa dk vkorZdky Kkr dhft;s % (a)ehVj iSekuk ftls 20 lseh fpUg ijyVdk;k x;k gSA (b) r f=kT;k o m nzO;eku okyh oy;] ftls bldh ifjf/k ij fLFkr fcUnq ls yVdk;k x;k gSA (c) aHkqtk okyh le:i oxkZdkj ifVV~dk ftls ,d dksus ls yVdk;k x;k gSA (d) m nzO;eku rFkk r f=kT;kokyh ,d pdrhftls dsUnz ls r/2 nwjh ij fLFkr fcUnq ls yVdk;k x;k gSA

48. (a) 1.51 s (b) 2 gr2

(c) 2 g3a8

(d) 2 g2r3

49. A uniform rod of length is suspended by an end and is made to undergo small oscillations. Find the lengthof the simple pendulum having the time period equal to that of the rod.yEckbZ dh ,d le:i NM+ dks ,d fljs ls yVdk dj vYi vk;ke ds nksyu djok;s tkrs gSA ml ljy yksyd dhyEckbZ Kkr dhft;sA ftldk vkorZdky NM+ ds vkorZdky ds rqY; gSA49. 2/3

50. A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimumpossible time period of the disc for small oscillations. What should be the distance of the hole from thecentre for it to have minimum time period ?r f=kT;k dh ,d le:i pdrh esa ,d fNnz djds] blls pdrh dks yVdk;k tkuk gSA vYi vk;ke ds nksyuksa ds fy;spdrh dk U;wure lEHko vkorZdky Kkr dhft;sA U;wure vkorZdky ds fy;s fNnz dh dsUnz ls nqjh fdruh gksxh\

50. 2 g2r

, r/ 2

51. A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time periodof oscillation of this pendulum. How does it differ from the time period calculated using the formula for asimple pendulum ?yksyd cukus ds fy;s 2 lseh f=kT;k dk ,d [kks[kyk xksyk] 18lseh yEcs /kkxs ls tksMk x;k gSA bl yksyd dk vkorZdkyKkr dhft;sA ;g ljy yksyd dk vkorZdky ds lw=k dh lgk;rk ls dh xbZ x.kuk ls fdl izdkj fuEu gksxk \

51. 0.89 s, it is about 0.3% larger than the calculated value

52. A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2 and time period2s. Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point ofsuspnsion as it goes through its mean position, (c) the acceleration of this particle as it goes through itsmean position and (d) the acceleration of this particle when it is at an exetreme position. Take g = 2 m/s2.nhokj ij yxh gqbZ dhy ls yVdk;k x;k ,d cUn o`kkdkj rkj 2vk;ke rFkk 2 ls.vkorZdky ds nksyu dj jgk gSAKkr dhft;s &(a) o`kkdkj rkj dh f=kT;k (b)tc ;g ek/; fLFkfr ls xqtjrk gS] rks fuyEcu fcUnq ls lokZf/kd nwjh okysd.k dh pky (c) bl d.k dk Roj.k tc ;g ek/; fLFkfr ls xqtjrk gS rFkk (d) tc ;g pje fLFkfr esa gksrk gS rks bld.k dk Roj.kA g = 2 m/s2 eku yhft;sA52. (a) 50 cm (b) 11 cm/s

(c) 1.2 cm/s2 towards the point of suspension (d) 34 cm/s2 towards the mean position

53. a uniform disc of mass m and radius r is suspended through a wire attached to its centre. If the time periodof the torsional oscillations be T, what is the torsional constant of the wire.m nzO;eku o r f=kT;k okyh ,d le:i pdrh] blds dsUnz ls tqM+s gq, rkj ls yVdk;h x;h gS A ;fn bl ejksM+h nksyddk vkorZdky T gS] rks rkj dk ,aBu fdruk gS\



53. 222

Tmr2

54. Two small balls, each of mass m are connected by a light rigid rod of length L. The system is suspended fromits centre by a thin wire of torsional constant k. The rod is rotated about the wire through an angle 0 andreleased. Find the tension in the rod as the system passes through the mean position.nks NksVh xSans] izR;sd dk nzO;ekum gS] LyEckbZ dh ,d gYdh ,oa n`< NM+ ds fljs tqMs gq, gSA bl fudk; dks dsUnzls] k ,aBu okys rkj ls yVdk;k x;k gSA NM+ dks rkj ds ikfjr% 0dks.k ls ?kqekdj NksM+ fn;k tkrk gSA tc fudk;ek/; fLFkfr ls xqtjrk gS] NM+ esa ruko Kkr dhft;sA

54.

2/122

2

40

2gm

Lk

55. A particle is subjected to two simple harmonic motions of same time period in the same direction. Theamplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if thephase difference between the motions is (a) 0, (b) 60, (c) 90.,d d.k ij ,d dh fn'kk esa leku vkorZdky okyh nks ljy vkorZ xfr;k vkjksfir dh tkrh gSA izFke xfr dkvkorZdky 3.0 lseh rFkk nqljh dk 4.0 lseh gSA ifj.kkeh vk;ke Kkr dhft;s] ;fn xfr;ksa ds e/; dykUrj gS % (a) 0,(b) 60, (c) 90.55. (a) 7.0 cm (b) 6.1 cm (c) 5.0 cm

56. Three simple harmonic motions of equal amplitudes Aand equal time periods in the same direction combine.The phase of the second motion is 60 ahead of the first and the phase of the third motion is 60 ahead of thesecond. Find the amplitude of the resultant motion.,d gh fn'kk esa leku vk;keArFkk leku vkorZdky okyh rhu vkorZ xfr;k la;ksftr dh tkrh gSA nwljh xfr izFkedh rqyuk esa dyk esa 60dks.k ls vkxs gS rFkk rhljh xfr] nwljh dh rqyuk esa dyk esa 60dks.k ls vkxs gSA ifj.kkehxfr dk vk;ke Kkr dhft;s56. 2 A

57. A particle is subjected to two simple harmonic motions given byx1 = 2.0 sin (100 t) and x2 = 2.0 sin (120 t + /3)

where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.,d ij vkjksfir nks ljy vkorZ xfr;k fuEukuqlkj xfr dk vk;ke Kkr dhft;s

x1 = 2.0 sin (100 t) rFkk x2 = 2.0 sin (120 t + /3)tgk x lseh esa rFkk t lsd.M esa gSA d.k dk foLFkkiu Kkr dhft;s (a) t = 0.0125, (b) t = 0.025.57. (a) 2.41 cm (b) 0.27 cm

58. A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line makingan angle of 45 with the X-axis. The two motions are given by

x = x0 sint and s = s0 sintFind the amplitude of the resultant motion.,d d.k ij nks ljy vkorZxfr;k vkjksfir dh tkrh gS] ,d X-v{k ds vuqfn'k rFkk nwljh X-v{k ls 45dks.k cukusokyh js[kk ds vuqfn'kA nksuks a xfr;k

x = x0 sint and s = s0 sint}kjk O;Dr dh tkrh gSA ifj.kkeh xfr dk vk;ke Kkr dhft;sA

58. 2/1002020 sx2sx

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Simple Harmonic Motion