Page No 360: - Chapter 14 Oscillations class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 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.

Answer:

(a) Time period, T = 2 s

Amplitude, A = 3 cm

At time, t = 0, the radius vector OP makes an angle   with the positive x-axis, i.e., phase angle

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given by the displacement equation:

(b) Time period, T = 4 s

Amplitude, a = 2 m

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = + π

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given as:

Question 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

Answer:

(a) 

If this equation is compared with the standard SHM equation , then we get:

The motion of the particle can be plotted as shown in the following figure.

(b) 

If this equation is compared with the standard SHM equation , then we get:

The motion of the particle can be plotted as shown in the following figure.

(c) 

If this equation is compared with the standard SHM equation , then we get:

Amplitude, A = 3 cm

Phase angle,   = 135°

Angular velocity,

The motion of the particle can be plotted as shown in the following figure.

(d) x = 2 cos πt

If this equation is compared with the standard SHM equation , then we get:

Amplitude, A = 2 cm

Phase angle, Φ = 0

Angular velocity, ω = π rad/s

The motion of the particle can be plotted as shown in the following figure.

Question 14.13:

Figure 14.30 (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 14.30 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

(a) What is the maximum extension of the spring 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?

Answer:

(a) For the one block system:

When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:

F = kl

Where, k is the spring constant

Hence, the maximum extension produced in the spring, 

For the two block system:

The displacement (x) produced in this case is:

Net force, F = +2 kx 

(b) For the one block system:

For mass (m) of the block, force is written as:

Where, x is the displacement of the block in time t

It is negative because the direction of elastic force is opposite to the direction of displacement.

Where,

ω is angular frequency of the oscillation

∴Time period of the oscillation, 

For the two block system:

It is negative because the direction of elastic force is opposite to the direction of displacement.

Where,

Angular frequency, 

∴Time period,