CBSE NCERT Solutions for Class 11 Physics Chapter 1 · 2020. 12. 17. · Class- XI-CBSE-Physics Oscillations Practice more on Oscillations Page - 1 CBSE NCERT Solutions for Class

Class- XI-CBSE-Physics Oscillations

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CBSE NCERT Solutions for Class 11 Physics Chapter 1

Back of Chapter Questions

14.1. Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other

and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its center of mass.

(d) An arrow released from a bow.

Solution: (b) and (c)

(a) The motion of the swimmer is not periodic. The swimmer's movement between

a river's banks doesn't have a definite period. As the swimmer's time during his

trip back and forth may not be the same.

(b) A freely suspended magnet's motion, if displaced and released from its N-S

direction, is periodic. This is because the magnet will oscillate with a definite

period of time about its position.

(c) When a molecule of hydrogen rotates around its centre of mass, it comes to the

same position again and again after an equal interval of time. Such motion is

periodic.

(d) The arrow released from the bow moves in the forward direction only. It's not

coming back. Hence, this motion is not periodic.

14.2. Which of the following examples represent (nearly) simple harmonic motion and which

represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a

point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Solution: (b) and (c) are SHMs

(a) and (d) are periodic, but not SHMs

(a) During its rotation about its axis, the earth comes to the same position again and

again in equal intervals of time. Hence, it is a periodic motion. However, this

motion is not simple harmonic. This is because the earth does not have a to and

fro motion about its axis.

(b) An oscillating mercury column in a U-tube is simple harmonic motion. This is

because the mercury moves to and fro on the same path, about the fixed position,

with a certain period of time.

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(c) When the ball is released inside a curved smooth bowl, it moves to and fro about

the lowermost point of the bowl with the same period of time, again and again.

Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation and its

vibration is the superposition of individual simple harmonic motions. Hence, it

is not simple harmonic, but periodic.

14.3. Figure 14.27 depicts four x-t plots for linear motion of a particle. Which of the plots

represent periodic motion? What is the period of motion (in case of periodic motion)?

Solution:

(b) and (d) are periodic

(a) It's not a periodic motion. This is a unidirectional motion and there is no

repetition of motion.

(b) The motion of the particle repeats itself after 2 s. Hence, it is a periodic motion,

with a time period of 2 s.

(c) It is not a periodic motion. This is because the particle does not repeat its motion.

For periodic motion, the entire motion of the particle must be repeated in equal

intervals of time.




(d) After 2 s, the particle's motion repeats itself. Hence, it is a periodic motion with

a time period of 2 s.

14.4. Which of the following functions of time represent (a) simple harmonic, (b) periodic

but not simple harmonic, and (c) non-periodic motion? Give period for each case of

periodic motion (ω is any positive constant):

sin ωt – cos ωt

sin 3ωt

3 cos (π

4– 2ωt)

cos ωt + cos 3ωt + cos 5ωt

exp(– ω2 t2)

1 + ωt + ω2 t2

Solution:

The given function represents a simple harmonic motion.

The given function is:

sin ωt − cos ωt

= √2 [1

√2 sin ωt −

1

√2 cos ωt]

= √2 [sin ωt × cosπ

4− cos ωt × sin

π

4 ]

= √2 sin (ωt −π

4)

As it can be written as a sin(ωt + ϕ), it represents simple harmonic motion

Its period is: 2π/ω

The given function represents a periodic, but not a simple harmonic motion.

The given function is:

sin 3ωt = 3 sin ωt – 4 sin3 ωt

Even though the two sin ωt represent simple harmonic motions respectively, but they

are periodic because superposition of two simple harmonic motion is not simple

harmonic.

The given function represents a simple harmonic motion.

3 cos (π

4– 2ωt) = 3 cos (2ωt –

π

4)

As it can be written as a cos(ωt + ϕ), it represents simple harmonic motion

Its period is: π/ω

The given function represents a periodic, but not a simple harmonic motion.




In cos ωt + cos 3ωt + cos 5ωt, each cosine function represents simple harmonic

motion, but the superposition of simple harmonic motion gives periodic.

As it is an exponential function, which is non-periodic as it does not repeat itself.

1 + ωt + ω2 t2 is a polynomial function which is non-periodic.

14.5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm

apart. Take the direction from A to B as the positive direction and give the signs of

velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and

(f) at 4 cm away from B going towards A.

Solution:

(a) Zero, Positive, Positive

(b) Zero, Negative, Negative

(c) Negative, Zero, Zero

(d) Negative, Negative, Negative

(e) Positive, Positive, Positive

(f) Negative, Negative, Negative

(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity

is zero, acceleration is positive as it is directed along AO and force is also

positive.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity

is zero, acceleration is negative as it is directed along BO and force is also

negative.




(c) The particle is executing a simple harmonic motion. O is the mean position of

the particle. Its velocity at the mean position O is the maximum. The value for

velocity is negative as the particle is directed leftward. The acceleration and

force of a particle executing SHM is zero at the mean position.

(d) The particle is at point Q and moving toward point O from the end B. This

direction of motion is from B to A. Hence, the particle's velocity and

acceleration, and the force on it are all negative.

(e) The particle is at point R and moving toward point O from the end A. This

direction of motion is from A to B, which is the conventional positive direction.

Hence, the values for velocity, acceleration, and force are all positive.

(f) The particle is at point S and moving toward point O from the end B. This

direction of motion is from B to A. Hence, the particle's velocity and

acceleration, and the force on it are all negative.




14.6. Which of the following relationships between the acceleration a and the displacement

x of a particle involve simple harmonic motion?

(a) a = 0.7x

(b) a = – 200x2

(c) a = – 10x

(d) a = 100x3

Solution: (c)

A motion represents simple harmonic motion if it is governed by the force law:

F = −kx

ma = −k

∴ a = −kx / m

Where F is the force

m is the mass (a constant for a body)

x is the displacement from the mean position

a is the acceleration

k is a constant

Among the given equations, only equation a = −10x is written in the above form with

k/m = 10.

Hence, this relation represents SHM.

14.7. The motion of a particle executing simple harmonic motion is described by the

displacement function, x(t) = A cos(ωt + ϕ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s,

what are its amplitude and initial phase angle? The angular frequency of the particle is

π s–1. If instead of the cosine function, we choose the sine function to describe the SHM

: x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the

above initial conditions.

Solution:

Given,

Initially, at t = 0




Displacement, x = 1 cm

Initial velocity, v = ω cm/sec.

Angular frequency, ω = π rad/s

It is given that:

x(t) = A cos(ωt + ϕ) . . . (i)

1 = A cos( ωx0 + ϕ) = A cos ϕ

A cos ϕ = 1 . . . (ii)

Velocity, v = dx/dt

differentiating equation (i) w.r.t ‘t’

v = – Aω sin( ωt + ϕ)

Now at t = 0; v = ω and

⇒ ω = – Aω sin (ωt + ϕ)

⇒ 1 = – A sin(ω × 0 + ϕ) = −A sin(ϕ)

⇒ A sin(ϕ) =– 1 . . . (iii)

Adding and squaring equations (ii) and (iii), we get:

A2(sin2 ϕ + cos2 ϕ) = 1 + 1

⇒ A = √2

The amplitude of oscillation is √2 cm.

Dividing equation (iii) by (ii), we get:

tan ϕ = −1

Thus, the initial phase angle, ϕ = 3π/4 , 7π/4

Now if simple harmonic motion is given as:

x = B sin(ωt + α)

Put the values in the above equation,

1 = B sin(ωx0 + α)

⇒ B sin α = 1 . . . (iv)

Also, velocity (v) = ω B cos(ωt + α)

Substituting the values, we get:

π = π B sin α

⇒ B sin α = 1 . . . (v)

Squaring and adding the equations (iv) and (v), we will get:




B2[sin2 α + cos2 α] = 2

Therefore, B = √2

Dividing equation (iv) by equation (v), we will get:

B sin α

B cos α= 1

tan α = 1 = tan(π/4)

Therefore, α = π/4 , 5π/4, . . . . ..

14.8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20

cm. A body suspended from this balance, when displaced and released, oscillates with

a period of 0.6 s. What is the weight of the body?

Solution:

Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg

Where,

g = acceleration due to gravity = 9.8 m/s2

F = 50 × 9.8 = 490

∴Spring constant, k =F

l=

490

0.2= 2450 N m−1

The mass m is suspended from the balance.

Time period, T = 2π√m

k

∴ m = (T

2π)

2

× k = (0.6

2 × 3.14)

2

× 2450 = 22.36 kg

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N

14.9. A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as

shown in Fig. 14.28. A mass of 3 kg is attached to the free end of the spring. The mass

is then pulled sideways to a distance of 2.0 cm and released.




Determine

(i) the frequency of oscillations,

(ii) maximum acceleration of the mass, and

(iii) the maximum speed of the mass.

Solution:

Given:

Spring constant, k = 1200 N/m

Mass, m = 6 kg

Displacement, A = 4.0 cm = 0.04 cm

Frequency of oscillation, f =1

T=

1

2π √

k

m

Where, T = time period.

Therefore f =1

2× 3.14 √

1200

3 = 3.18 Hz

Maximum acceleration (a) = ω2 A

Where, ω = Angular frequency = √k

m

A = maximum displacement

Therefore, a = A(k/m)

a = 0.02 × (1200

3) = 8 m /s−2

Maximum velocity, Vmax = ω A

= 0.02 × √1200

3

Therefore, VMAX = 0.4 m/s

Hence, the maximum velocity of the mass is 0.4 m/s.

14.10. In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a

function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is

(a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.




In what way do these functions for SHM differ from each other, in frequency, in

amplitude or the initial phase?

Solution:

Given:

Spring constant, k = 1200 N/m

Mass, m = 6 kg

Displacement, A = 4.0 cm = 0.04 cm

Angular frequency, ω = √k

m= √

1200

6= 14.14 s−1

(a) Since time is measured from mean position,

x = A sin ω t

x = 4 sin 14.14t

At the maximum stretched position, the mass has an initial phase of π/2 rad.

Then, x = A sin ( ωt +π

2) = A cos ωt = 4 cos 14.14t

(b) At the maximum compressed position, the mass is at its leftmost position with

an initial phase of 3π/2 rad.

Then, x = A sin ( ωt +3π

2) = −4 cos14.14 t

14.11. Figures 14.29 correspond to two circular motions. The radius of the circle, the period

of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-

clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius

vector of the revolving particle P, in each case.

Solution:

For time period, T = 2 s

Amplitude, A = 3 cm




At the time, t = 0, the radius vector OP makes an angle π

2 with the positive x-axis, i.e.,

Phase angle ϕ =π

2

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time

t is:

x = A cos [2πt

T+ ϕ]

= 3 cos [2πt

2+

π

2]

= −3sin (πt )

= −3sin (πt ) cm

Time period, T = 4 s

Amplitude, A = 2 m

At time t = 0, OB makes an angle π with the x-axis, in the anticlockwise direction.

Thus, phase angle, ϕ = + π

Therefore, the equation of simple harmonic motion for the x-projection of OB, at time

t is:

x = A cos [2πt

T+ ϕ]

= 2 cos [2πt

4+ π ]

= −2 cos (πt

2) cm

14.12. Plot the corresponding reference circle for each of the following simple harmonic

motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and

the angular speed of the rotating particle. For simplicity, the sense of rotation may be

fixed to be anticlockwise in every case: (x is in cm and t is in s).

(a) x = – 2 sin (3t +π

3)

(b) x = cos (π

6– t)

(c) x = 3 sin (2πt +π

4)

(d) x = 2 cos πt

Solution:

(a) x = – 2 sin (3t +π

3)

= 2 cos (3t +π

3+

π

2)




= 2 cos (3t +5π

6)

Compare this equation to standard equation, x = A cos(ωt + θ)

Amplitude, A = 2 cm

Angular frequency, ω = 3 rad/s

Phase angle, θ = 5π/6

See the figure here we graph of it.

(b) x = cos (π

6– t) = cos (t −

π

6)


Amplitude, A = 1 cm

Angular frequency, ω = 1 rad/s

Phase angle, θ = −π/6


(c) x = 3 sin(2πt + π/4)




= − 3 cos ( 2πt +π

4+

π

2) [cos (

π

2+ θ) = − sin θ]

= −3 cos(2πt + 3π/4)

= 3 cos( 2πt + 3π/4 + π)

= 3 cos(2πt + 7π/4)


Amplitude, A = 3 cm

Angular frequency, ω = 2π rad/s

Phase angle, θ = 7π/4


(d) x = 2 cos πt


Amplitude, A = 2cm

Angular frequency, ω = π rad/s

Phase angle, θ = 0





14.13. Figure (a) shows a spring of force constant k clamped rigidly at one end and a

mass m attached to its free end. A force F applied at the free end stretches the

spring. Figure (b) shows the same spring with both ends free and attached to a

mass m at either end. Each end of the spring in figure (b) is stretched by the

same force F.

(a) What is the extension in the spring at equilibrium in the two cases?

(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the

period of oscillation in each case?

Solution:

In both cases at equilibrium, the external force is balanced by the spring force.

F = kx

x =F

k

Thus, the extension produced in the spring, x =F

k

For one block system:

Force on the block of mass m is due to the spring, F = −kx

From Newton's 2nd law, F = ma

Where x is the displacement of the block in time t




Therefore, ma = −kx

The negative sign is present as the direction of the elastic force is opposite to the

direction of the displacement.

Compare with a = −ω²x, then we will get

ω2 = k/m

ω is the angular frequency of oscillation.

Therefore, the time period of oscillation,

T =2π

ω= 2π√

m

k

For two block system:

The system is divided into two similar systems with spring divided in two equal halves.

k′ = 2k

So, F = −k′x = −2kx

From Newton's 2nd law,

F = ma

So, − 2Kx = ma

a = − (2K/m)x

Compare with a = −ω²x

ω2 = 2k/m

ω is the angular frequency of oscillation.

Therefore, the time period of oscillation,

T =2π

ω= 2π√

m

2k

14.14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of

1.0 m. If the piston moves with simple harmonic motion with an angular frequency of

200 rad/min, what is its maximum speed?

Solution:

Given:

The angular frequency of the piston, ω = 200 rad/ min.

Stroke = 1.0 m

Amplitude, A = 1/2 = 0.5 m

Thus, maximum speed Vmax = Aω = 100 m/ min




14.15. The acceleration due to gravity on the surface of moon is 1.7 m s−2. What is the time

period of a simple pendulum on the surface of moon if its time period on the surface of

earth is 3.5 s? (g on the surface of earth is 9.8 m s−2)

Solution:

Given:

Acceleration due to gravity on the surface of the moon, g’ = 1.7 m s−2

Acceleration due to gravity on the surface of the earth, g = 9.8 m s−2

The time period of a simple pendulum on earth, T = 7s

We know, the time period of a simple pendulum is given as

T=2π√l

g

Where l is the length of the pendulum l=T2

(2π)2 × 9.8 = 12.17m

On the moon’s surface, time period,

T = 2π√l

g′

Therefore,

T = 2π√12.17

1.7= 16.82 s

14.16. Answer the following questions:

Time period of a particle in SHM depends on the force constant k and mass m of the

particle:

T = 2π√m

k. A simple pendulum executes SHM approximately. Why then is the time

period of a pendulum independent of the mass of the pendulum?

The motion of a simple pendulum is approximately simple harmonic for small angle

oscillations. For larger angles of oscillation, a more involved analysis shows that T is

greater than 2π√l

g. Think of a qualitative argument to appreciate this result.

A man with a wristwatch on his hand falls from the top of a tower. Does the watch give

correct time during the free fall?

What is the frequency of oscillation of a simple pendulum mounted in a cabin that is

freely falling under gravity?

Solution:




The time period of a simple pendulum, T = 2π√m

k

For a simple pendulum, the force constant k depends on the mass of the particle. Hence

their ratio becomes independent of mass.

Thus, the time period T, of a simple pendulum is independent of its mass. The time

period of a simple pendulum is given by the formula, 2π√l

g

In the case of a simple pendulum, the restoring force acting on the bob of the pendulum

is:

F = – mg sinθ

Where,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ ≈ θ

For large θ, this approximation is no longer valid and the value of sin θ is greater than

θ. This decreases the effective value of g.

Thus, the time period increases as 2π√l

geff

The wristwatches do not have a simple pendulum to measure time. Hence, working off

a wristwatch does not depend on the acceleration due to gravity. So, the time shown by

it will be correct during free fall.

During free fall the effective acceleration due to gravity is zero on the object. Hence

the pendulum will not oscillate and its frequency will be zero.

14.17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The

car is moving on a circular track of radius R with a uniform speed v. If the pendulum

makes small oscillations in a radial direction about its equilibrium position, what will

be its time period?

Solution:

The bob of the simple pendulum will experience centripetal acceleration because by the

circular motion of the car and the acceleration due to gravity.

Acceleration due to gravity = g

Centripetal acceleration =v2

R

Where

v is the uniform speed of the car




R is the radius of the track

From the figure, we can see the effective acceleration (aeff ) is given as:

aeff = √g2 + (v2

R)

2

The time period, T=2π√l

aeff

Where l is the length of the pendulum

Therefore,

Time period, T = 2π√l

√g2+v4

R2

14.18. A cylindrical piece of cork of density of base area A and height h floats in a liquid of

density ρl. The cork is depressed slightly and then released. Show that the cork

oscillates up and down simple harmonically with a period

T = 2π√hρ

ρlg

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Solution:

Given:

The base area of the cork = A

Height of the cork = h

The density of the liquid = ρl

The density of the cork = ρ

In equilibrium:




The weight of the cork = Weight of the liquid displaced by the floating cork.

Let the cork be dipped slightly by y as shown in the figure.

As a result, some excess water of a certain volume is displaced. Thus, an extra up-thrust

provides the restoring force to the cork.

Up-thrust = Restoring force,

F = Weight of the extra water displaced

F = – (Volume × Density × g)

[∴ Volume = Area × Distance through which the cork is depressed = Ay]

F = – Ayρlg . . . (i)

Now,

F = ky

k = F/y

Where k is a constant and can be written as

k = F/y = – A x ρl g . . . (ii)

The time period of oscillations of the cork:

T = 2π√m

k

Where,

m = mass of the cork

= Volume × density

= Base area of the cork × Height of the cork × Density of the cork

= Ahρ

Now, the time period becomes:




T=2π√hρ

ρlg

14.19. One end of a U-tube containing mercury is connected to a suction pump and the other

end to atmosphere. A small pressure difference is maintained between the two columns.

Show that, when the suction pump is removed, the column of mercury in the U-tube

executes simple harmonic motion.

Solution:

When the suction pump is removed the mercury level on one side will be raised

compared to the side as shown in the figure.

Area of the cross-section of the U-tube = A

The density of the mercury column = ρ

Acceleration due to gravity = g

Restoring force, F = Weight of the column of mercury of a certain height

F = – (Volume × Density × g)

F = – (A × 2y × ρ × g)

= – 2Aρgy

= – k × Displacement in one arm (h)

Where 2y is the height of the mercury column in the two arms and k is a constant.

The value of k is given by,

k = −F

y= 2Aρg

Now the time period of oscillation can be given by,

T=2π√m

k= 2π√

m

2Aρg

Where, m, is the mass of the mercury column




Let, l be the length of the total mercury in the U-tube.

Mass of mercury, m = Volume of mercury × Density of mercury = Alρ

Therefore,

T=2π√Alρ

2Aρg⇒T=2π√

l

2g

If we consider the length of the mercury column is very high then, l = 2 and the time

period of oscillation can be written as,

T=2π√2h

2g= 2π√

h

g

Additional Exercises

14.20. An air chamber of volume V has a neck area of cross section a into which a ball of mass

m just fits and can move up and down without any friction (Fig.14.33). Show that when

the ball is pressed down a little and released, it executes SHM. Obtain an expression

for the time period of oscillations assuming pressure-volume variations of air to be

isothermal [see Fig. 14.33].

Solution:

Let the ball is depressed by x units. As a result, the volume will decrease, and this would

increase the pressure inside.

decrease the volume of air (∆V) = cross-section area of neck(a) × depression length = ax

∆V = ax

Volume strain =∆V

V =

ax

V

We know the bulk modulus is given as, B = −P

∆V

V

[∴volume stress = bulk stress = P]

We can write,




B = −P

ax/V

P = −Bax

V

Also, we know, F/a = P [ pressure = force/area]

F

a= −

Bax

V

F = −Ba2x

V

Compare with F = −mω²x

ω = √Ba2

mV

T = 2π√mV

Ba2

14.21. You are riding in an automobile of mass 3000 kg. Assuming that you are examining the

oscillation characteristics of its suspension system. The suspension sags 15 cm when

the entire automobile is placed on it. Also, the amplitude of oscillation decreases by

50% during one complete oscillation. Estimate the values of

(a) the spring constant k and

(b) the damping constant b for the spring and shock absorber system of one wheel,

assuming that each wheel supports 750 kg.

Solution:

Mass of the automobile, m = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

The equation for the restoring force for the system:

F = – 4kx = mg

Where k is the spring constant of the suspension system.

Time period, T = 2π√m

4k

And k =mg

4x=

3000×10

4×0.15= 50000 N/m

Spring constant, k = 5 × 104 N/m

Each wheel supports a mass, M =3000

4= 750 kg




For damping factor b, the equation for displacement is written as: x = x0e−bt

2M

The amplitude of oscillation decreases by 50%.

∴ x =x0

2

x0

2= x0e−

bt2M ⇒

1

2= e−

bt2M

Take natural log on both side

ln 2 =bt

2M⇒ b =

2M ln 2

t

Where t = 2π√m

4k= 2π√

3000

4×5×104= 0.7691 s

∴ b =2M ln 2

t=

2 × 750 × 0.693

0.7691= 1351.58 kg/s

Therefore, the damping constant of the spring is 1351.58 kg/s.

14.22. Show that for a particle in linear SHM the average kinetic energy over a period of

oscillation equals the average potential energy over the same period.

Solution:

The equation of displacement of a particle executing SHM at an instant t is given as:

x = A sin ωt

Where, A = Amplitude of oscillation

ω = Angular frequency

The velocity of the particle is:

v =dx

dt= Aω cos ωt

The kinetic energy of the particle is:

Ek =1

2Mv2 =

1

2M(Aω cos ωt)2

The potential energy of the particle is:

Ep =1

2kx2 =

1

2M(A sin ωt)2

For time period T, the average kinetic energy over a single cycle is given as:

(Ek)avg =1

T∫

1

2MA2ω2 cos2 ωt dt

T

0

=1

2TMA2ω2 ∫ cos2 ωt dt

T

0




=1

2TMA2ω2 ∫

cos 2ωt + 1

2dt

T

0

=1

2TMA2ω2 [

sin 2ωt

2ω+ t]

0

T

=1

4TMA2ω2T =

1

4MA2ω2 … (i)

And, average potential energy over one cycle is given as:

(Ep)avg

=1

T∫

1

2MA2ω2 sin2 ωt dt

T

0

=1

2TMA2ω2 ∫ sin2 ωt dt

T

0

=1

2TMA2ω2 ∫

1 − cos 2ωt

2dt

T

0

=1

2TMA2ω2 [t −

sin 2ωt

2ω]

0

T

=1

4TMA2ω2T =

1

4MA2ω2 … (ii)

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given

time period is equal to the average potential energy for the same time period.

14.23. A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is

twisted by rotating the disc and released. The period of torsional oscillations is found

to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of

the wire. (Torsional spring constant α is defined by the relation J =– α θ, where J is the

restoring couple and θ the angle of twist).

Solution:

Given:

Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The time period of torsional oscillations, T = 1.5 s

∴ The angular frequency, ω =2π

T=

2π

1.5 rad/s

The moment of inertia of the disc is:

I =1

2 mr2 =

1

2× 10 × (0.15)2 = 0.1125 kg m2

The restoring couple on disc is given as, J =– α θ

The acceleration of disc can be written as, J

I=–

(α θ)

0.1125




We know in angular SHM, the angular acceleration = −ω2θ

Compare it with the angular acceleration of disc.

(α θ)

0.1125= −ω2θ

⇒ ω2 =α

0.1125

Now,

(2π

1.5)

2

=α

0.1125

α = 1.972 N m/rad

Hence, the torsional spring constant of the wire is 1.972 N m/rad.

14.24. A body describes simple harmonic motion with an amplitude of 5 cm and a period of

0.2 s. Find the acceleration and velocity of the body when the displacement is

(a) 5 cm,

(b) 3 cm,

(c) 0 cm.

Solution:

Given:

Amplitude, A = 5 cm = 0.05 m

Time period, T = 0.2 s

∴ The angular frequency ω =2π

T= 10π rad/s

For displacement x,

The velocity of the particle is given as, v = ω√A2 − x2

The acceleration of the particle is given as, a = −ω2x

x = 5 cm = 0.05 m

The velocity, v = 10π√0.052 − 0.052 = 0 m/s

The acceleration, a = −(10π)2(0.05) = −5π2 m/s2

x = 3 cm = 0.03 m

The velocity, v = 10π√0.052 − 0.032 = 0.4π m/s

The acceleration, a = −(10π)2(0.03) = −3π2 m/s2

x = 0 cm = 0 m

The velocity, v = 10π√0.052 − 02 = 0.5π m/s




The acceleration, a = −(10π)2(0) = 0 m/s2

14.25. A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal

plane without friction or damping. It is pulled to a distance x0 and pushed towards the

centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting

oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x =a cos (ωt + θ) and note that the initial velocity is negative.]

Solution:

The displacement equation for an oscillating mass is given by:

x = A cos(ωt + θ)

Where,

A is the amplitude

x is the displacement

θ is the phase constant

Velocity, v =dx

dt = Aω sin(ωt + θ)

At t = 0, x = x0

x0 = A cos θ … (i)

And,

v0 = Aω sin θ

v0

ω= A sin θ … (ii)

Squaring and adding equations (i) and (ii), we get:

A2(cos2 θ + sin2 θ) = x02 + (

v0

ω)

2

∴ A = √x02 + (

v0

ω)

2

Hence, the amplitude of the resulting oscillation is √x02 + (

v0

ω)

2

⧫ ⧫ ⧫


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