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

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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}

A=4.9cm

Solution 3

{A=amplitude}

K.E = P.E

A=10cm

Particle distance from mean position=cm

Solution 4

A -(1)

-(2)

{a=acceleration, A=amplitude}

From (1),

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

cm from mean position.

Solution 5

Compare it with,

(a) Amplitude(A)=2.0cm

Time periodsec

For spring constant,

(b) At t=0,

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 1^st time when

It becomes '0' when

sec

(b) Acceleration,

It is having negative sign. To have maximum acceleration,

On solving, sec

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

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

Solution 12

Given, m=2kg, T=4sec

Time period,

N/m

Now,

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) =massacceleration

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

(b)

For small block

Normal contact will be minimum at highest point.

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)

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.

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

sec

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

(e)

PE=4.5J

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

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.

Solution 18

(a)

Maximum displacement is given by

(b) Energy stored at equilibrium=

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,

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'

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,

U=0

sec

For right incline

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

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

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=

Solution 42

(b) Force along tunnel=

Force perpendicular to tunnel=

(d) Resultant force is

(e) mass acceleration

(towards center)

So, body executes SHM.

Solution 43

(a) Restoring force

(b)

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)

(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)

(b)

(c)

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

(b) t=0.025sec

Solution 58

Phase difference,

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

Simple Harmonic Motion Exercise 252

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Simple Harmonic Motion Exercise 253

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Simple Harmonic Motion Exercise 254

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Simple Harmonic Motion Exercise 255

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Simple Harmonic Motion Exercise 256

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

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