Page No 362: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 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.

Answer:

(a) 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, 

And   = 5000 = 5 × 10⁴ N/m

Spring constant, k = 5 × 10⁴ N/m

(b) Each wheel supports a mass, M =  = 750 kg

For damping factor b, the equation for displacement is written as:

The amplitude of oscillation decreases by 50%.

∴

Where,

Time period,   = 0.7691 s

= 1351.58 kg/s

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

Question 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.

Answer:

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

Where,

A = Amplitude of oscillation

ω = Angular frequency 

The velocity of the particle is:

The kinetic energy of the particle is:

The potential energy of the particle is:

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

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

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.

Question 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).

Answer:

Mass of the circular disc, m = 10 kg

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

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

I 

=  × (10) × (0.15)²

= 0.1125 kg m²

Time period, 

α is the torsional constant.

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad^–1.

Question 14.24:

A body describes simple harmonic motion with 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.

Answer:

Amplitude, A = 5 cm = 0.05 m

Time period, T = 0.2 s

(a) For displacement, x = 5 cm = 0.05 m

Acceleration is given by:

a = –

Velocity is given by:

v

= 0

When the displacement of the body is 5 cm, its acceleration is –5π² m/s² and velocity is 0.

(b) For displacement, x = 3 cm = 0.03 m

Acceleration is given by:

a =

Velocity is given by:

v

= 0.4 π m/s

When the displacement of the body is 3 cm, its acceleration is –3π m/s² and velocity is 0.4π m/s.

(c) For displacement, x = 0

Acceleration is given by:

a

= 0

Velocity is given by:

When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5 π m/s.

Question 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.]

Answer:

The displacement equation for an oscillating mass is given by:

x = 

Where,

A is the amplitude

x is the displacement

θ is the phase constant

Velocity,

At t = 0, x = x₀

x₀ = Acosθ = x₀ … (i)

And,

 … (ii)

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

Hence, the amplitude of the resulting oscillation is .