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Class 11-science H C VERMA Solutions Physics Chapter 12 - Simple Harmonic Motion

Simple Harmonic Motion Exercise 252

Solution 1

S.H.M equation is

    -(1)

Amplitude, A=10cm

At t=0, position x=5cm

Time period=6sec

   

Substitute in eq (1)

     

    

    

Now, displacement equation is

    

    

At t=4sec

    

On solving    

Acceleration, a=   

    

    

Solution 2

Taking only magnitude, a=  

   

Acceleration, a=10m/  

Position, x=2cm=0.02m

   

Time period, T= sec

Velocity,   {A=amplitude}

   

   

  m

 A=4.9cm

Solution 3

  {A=amplitude}

  

K.E = P.E

   

   

   

 A=10cm

  

 cm

Particle distance from mean position= cm

Solution 4

 A -(1)

    -(2)

{a=acceleration, A=amplitude}

  

   

   

From (1),   

   

  cm

Now, speed=8cm/s, position=?

  

  

  

  

 cm from mean position.

Solution 5

   

Compare it with,   

(a) Amplitude(A)=2.0cm   

 Time period sec

 For spring constant,   

   

  

  

(b) At t=0,

  cm

 Velocity= m/s

 Acceleration,   

   

   

   {considering only magnitude} 

Solution 6

Give SHM equation

     

(a) It comes to rest when velocity=0

 'v' becomes '0' at extreme position

 At extreme position,  cm

   

   

   

   

  seconds

(b) We know that,   

   

   

   

   

   

    

  seconds

(c)   

 'v' is maximum only when   

   

   

  sec

 Not possible because time won't be negative.

   

   

  sec  

Solution 7

    

   

 (a) We know that velocity(v)  

   

 The particle velocity becomes zero for 1st time when   

   

 It becomes '0' when   

   

  sec

  sec

(b) Acceleration,   

 It is having negative sign. To have maximum acceleration,

   

   

 On solving,  sec

(c) Particle again comes to rest when

   

   

   

 On solving,  s

Solution 8

Let the equation of SHM be

   with time period T

 For half-amplitude,   

   

   -(1)

For   

   

   -(2)

Now   

   

   

  sec   

Solution 9

Length of the pendulum clock=1m

 sec

Given, k=0.1N/m

Time periods are same.

 sec

  

On solving, m=0.01kg=10gms  

Solution 10

Given

Spring frequency=Pendulum frequency

  

  [ length of pendulum clock=1m]

  

  

  [ ]

 Extension=1m  

Solution 11

Given,

Time period=0.314sec

Mass=0.5kg

Amplitude=0.1m

We know,   

   

 On solving,  N/m

 F=kx

  N

Nor. Max. force=weight + force due to extension

  N

Solution 12

Given, m=2kg, T=4sec

Time period,   

   

   

  N/m

Now,   

  

  

  m

Now, P.E =  

Simple Harmonic Motion Exercise 253

Solution 13

  

  

 N/m

In 1sec, it completes 5 revolutions.

So, time taken for 1rotation= sec

  

  

On solving, m=160N/m  

Solution 14

(a)  =mass acceleration

   

   

 Here -ve sign shows that force is towards equilibrium position.

(b)

For small block

  

  

  

Normal contact will be minimum at highest point.

(c) For maximum amplitude, x=A, N=0

   

   

   

Solution 15

(a) Let compression in spring be x in equilibrium position.

  

  

  

(b) Let blocks lose contact at y distance from equilibrium position.

For   block

  

   

  

  

  

y=x (At natural length of spring it loses contact)

(c) Spring is further compressed by 2x for extreme position

 Total compression=x+2x=3x

 Now,

 Total energy conservation at extreme position and natural length of spring.

  

  

 

 

  

  

  

  

Solution 16

(a) Let spring is compressed by x at equilibrium position.

  

  

  

 cm

(b) Total energy=K.E+ P.E

   

   

  J

(c) Time period,   

  sec

(d) At the time of maximum compression (v=0) and let spring is compressed by (x+a)

   

  

  

  

  

  

  

  

  

 cm

(e)   

   

 PE=4.5J

(f) At right extreme, extension in spring will be

  m

   

  J

Since, the external force is acting on block the energy will not be conserved.

Solution 17

(a) The springs are parallel here.

Equivalent spring constant in parallel case is

   

   

(b) We can consider the springs here as parallel. Because, they are not in linear and continuous connection. If the block moves, deformation in one spring increases. So, the force increases.

     

   

(c) Springs are in series,

  

    

Solution 18

(a)   

   

Maximum displacement is given by   

(b) Energy stored at equilibrium=  

   

   

(c) At mean position,

 P.E.=0

   

   

Solution 19

 

Let spring C is compressed by x at any instant,

then extension in spring   

and in spring   

Net force on block

  

  

  

  

  

  

  

Solution 20

 

Let spring C is compressed by x at any instant, then extension in

Spring A   

Spring B=   

Net force on block

  

  

  

  

  

Solution 21

To find frequency and amplitude, first resolve it.

  are in series.

So,   

Now   are in parallel.

Total,    

Now, frequency   

Amplitude,   

Solution 22

We know   

For   spring,   

 For   spring,   

For   spring,   

Now   are in series

  

  

  

Simple Harmonic Motion Exercise 254

Solution 23

Initially there will be some tension in spring.

  

After applying some force 'F', tension becomes

  

Resultant tension=   

  

  

But   

  

  

  

acceleration=    

Time period   

  

Solution 24

 Let at any instant the velocity of the block is v and spring is stretched by x.

So,

Total energy=   

Total energy is constant

  

  

  

  

  

Comparing with   

  

    

Solution 25

 The velocity of center of mass is zero, so it will remain at rest during motion.

If left block stretches spring by x so, right block will also stretch spring by x.

Total energy=   

TE remains constant

  

    

  

  

Comparing with   

  

   

Solution 26

Consider the entire rectangular body mass at center of mass point.

As it displaced, it will move.

Here   gives the moment in actual direction we required.

Now,     

    

    

For small angles,   

    

    

    

So, this motion is SHM.

 

 

  

Solution 27

Total mass of the system, M=3+1=4kg

Time period of SHM,

    

   sec

So, frequency=  Hz

Let v be the velocity of the 1kg block at mean position

 KE=TE

    

    

 v=1m/s

Let v' be the velocity of combined blocks

    

    

    m/s

Now, KE'=TE'

    

    

   cm 

Solution 28

 For same mass and elastic collision, blocks will exchange velocity on collision.

Now, block A c omes to rest and block B moves and execute half SHM.

Now again block B collides and come to rest. Block A moves and execute periodic motion for distance 2L with speed v.

  

  

  

Total time=    

Solution 29

The maximum height attained by ball on both inclined will be same as no energy loss.

For left incline,

  

 cm

  

U=0

  

  

 sec

For right incline

  

 cm

  

U=0

  

  

  

 sec

Total time=   

 =  sec  

Solution 30

The above situation can be drawn as

Considering both blocks and spring as a system, there is no external resultant force on the system. Hence, center of mass will remain at rest. The mean positions of the two SHM 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,

  -(1)

As COM remains at rest;   -(2)

From (1) and (2)

  

Hence, the left block is   distance away from its mean position in the beginning of the motion. The force by the spring on this block at this instant is equal to the tension of spring i.e.,   .

  

acc=   

  Angular frequency   

    

Solution 31

 Let plate is displaced by x

 

 

Vertical translatory equilibrium

  

Torque about A for rotational equilibrium.

  

  

  

and   

    

  

  

  

  

Net force in horizontal direction=   

  

  

  

  

  

Solution 32

  

Given, T=2sec

  

  

l=1m   

Solution 33

  

This is similar to SHM equation.

  

  

T=2sec

  

  

 m

Simple Harmonic Motion Exercise 255

Solution 34

We know Time period of pendulum clock = 2sec

1 oscillation  2.00 sec

x oscillations  1 day=24  3600 sec

   

Give T=2.04 sec

In each oscillation, it is slower by 2.04-2.00=0.04 sec

For 1 day=   min

  The clock runs 28.8 min slower in one day.

Solution 35

  

  

  

  

  

  

  

Solution 36

(a)  sec

 Frequency=   

(b) When it is on moon

    

    

    

    

Solution 37

Change in K.E = change in P.E

    

  

Now,   

 due to weight

 due to oscillation (centrifugal force)

  

    

  

Now,   

  

  

  

Solution 38

Torque about center of concave surface

  

  (distance)

  

  

[   is very small  ]

  

   

Comparing with   

  

   

Solution 39

Let   be the angular velocity of the system about the point of suspension at any point.

Velocity of the ball rolling on a rough concave surface   is given by,

  

Also,   

where    is the rotational velocity of the sphere.

  -(1)

As total energy of a particle in SHM remains constant,

 constant

Substituting the values of   and   in the above equation, we get:

 constant

    

 constant

 constant

Taking derivative on both sides, we get:

  

  

  

  

    

 =constant

Therefore, the motion is S.H.M

  

Time period is given by,     

Solution 40

Acceleration due to gravity at a depth of 1600 mts is

   

    

  

Time period at depth,   

l=0.4m

 sec

Solution 41

Force on particle of mass m at a distance x from center is

  

  

  

  

  

(a)   

 At y=R;    

    

    

    

    

Displacement-time equation given by

    

Let   and   be the time taken by the particle to reach the positions P and Q.

  Phase at P <   and   Phase at Q<   

At point P; y=R

    

    

At point Q; y=R

    

    

So,   

  

(b) By energy conservation,

At height R above tunnel=At tunnel surface

  

  

  

  [  ]

So, time to cover tunnel length=   

(c) By energy conservation, the velocity of ball on reaching back again at projection point will be  .

So, time to cover tunnel length=   

Solution 42

 

(b) Force along tunnel=   

    

     

 Force perpendicular to tunnel=   

    

    

(c) Normal force exerted by wall,   

    

(d) Resultant force is   

    

(e)  mass  acceleration

     

    (towards center)

    

 So, body executes SHM.

    

    

     

Solution 43

(a) Restoring force

  

  

  

  

  

  

  

(b)

  

  

  

  

  

(c) uniform velocity means a=0

    

Solution 44

Time period of simple pendulum having acceleration 'a' upward

    

    

    

 Squaring,

 32+a=36

     

Solution 45

Here velocity is uniform.

So, acceleration =   

  

  -(1)

If car is moving with acceleration  ,

Then   

     -(2)

  

  

Solution 46

 

  

  

  

  

Solution 47

(a)      

   sec

(b) If she sits on a merry-go-round, she and her earrings will have centripetal acceleration.

  

  

  

    

Now,  sec   

Solution 48

(a)

  

    

    

    

T=1.52sec

(b)

  

    

    

(c)

  

    

    

    

(d)

  

    

    

     

Solution 49

Time period of rod

  

    

  

Now,

  

  

    

Solution 50

Let distance of hole from center of disc be x

  

    

    

For minimum time period,

 Minimize,   

Differentiate with respect to x

  

    

  

  

Now, put x in T

  

    

Solution 51

 

    

    

 sec

Time period by considering it as simple pendulum

  

   sec

% more=   

Solution 52

(a)     

  

  

  

 cm

(b) Total energy at extreme position=Total energy at mean position

  

  

    

  

  

 rad/s

Speed of lowest point   

    

    cm/sec

(c)   

    

    

(d) At extreme position, v=0

 Radial acceleration,   

 Tangential acceleration,

    

    

    

Solution 53

    

  

  

  

Simple Harmonic Motion Exercise 256

Solution 54

Torque when rod is at an angle   about the wire.

  

Work done in rotating from   to 0

  

  

  

Work=   

This work is in form of rotational kinetic energy.

  

  

  

    

Solution 55

 cm;    cm

    

(a)   

   cm

(b)   

    

    

   cm

(c)   

    

   cm

Solution 56

By superposition principle,

Resultant of   and   vector is=   

 = A along   vector

Now, resultant of   vector and   vector= A+A= 2A 

Solution 57

Resultant displacement

    

    

(a) t=0.0125sec

    

    

   cm

(b) t=0.025sec 

    

    

   cm

Solution 58

Phase difference,    

  

  

    

  

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