Question 50: Waves

Question

A block $$M$$ hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to fixed rigid support at $$O.$$ A transverse wave pulse (Pulse 1) of wavelength $${\lambda _0}$$ is produced at point $$O$$ on the rope. The pulse takes time $${T_{OA}}$$ to reach point $$A.$$ If the wave pulse of wavelength $${\lambda _0}$$ is produced at point $$A$$ (Pulse 2) without disturbing the position of $$M$$ it takes time $${T_{AO}}$$ to reach point $$O.$$ which of the following options is/are correct?

Accepted Answer

The velocity of the transverse wave is given by
$$v = \sqrt {{T \over \mu }} $$ ..... (i)
Where T is the tension in the rope and is given by
T = Mg + $$\mu$$yg
Where y = distance of point P on the rope from its bottom
L = total length of the rope
Put the value of T in eqn. (i), we get
$$v = \sqrt {{{Mg + \mu yg} \over \mu }} $$ ...... (ii)
Also, speed of pulse 1 and pulse 2 are same at each point because speed is a characteristic of the medium. Hence, speed of the two pulses are same at the mid point (x = l/2). However, since pulse 1 is moving in +x direction and pulse 2 is moving in Ã¢ÂÂx direction, their velocities (which is a vector quantity) at the mid point are not same.
So, option (b) is incorrect.
Since both the pulse travel in the same medium and velocity is medium dependent. The velocity of any pulse along the rope depends on tension in the rope and linear mass density so it is independent of its frequency(characteristic of the source of pulse generation) and wavelength.
So, option (d) is correct.
Also, v = $$
u  $$$$\lambda$$ or $$\lambda$$ $$\propto$$ v
($$\because$$ frequency of a wave depends on the source)
For pulse 1, the velocity decreases at point A as tension decreases. So the wavelength of pulse 1 becomes shorter when it reaches point A.
So, option (c) is incorrect.
From eqn. (ii), the velocity is given as
$$v = \sqrt {{{Mg + \mu yg} \over \mu }} $$ or $${{dy} \over {dt}} = \sqrt {{{Mg + \mu yg} \over \mu }}  \Rightarrow dt = {{dy} \over {\sqrt {{{Mg + \mu yg} \over \mu }} }}$$
Integrating both sides,
$$\int\limits_0^T {dt = \int\limits_0^L {{{dy} \over {\sqrt {{{Mg} \over \mu } + yg} }}} } $$; $$	herefore$$ $$T = {2 \over g}\left( {\sqrt {{{Mg} \over \mu } + Lg}  - \sqrt {{{Mg} \over \mu }} } ight)$$
Where T is time taken by pulse to travel distance L.
Since T depends upon g, M, L and $$\mu$$ which are constant.
$$	herefore$$ Time taken by pulse 1 to reach point A and time taken by pulse 2 to reach point O is same.
$$	herefore$$ T
AO
= T
OA
; So option (a) is correct.

Answer

The time $${T_{AO}} = {T_{OA}}$$

Answer

The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope

Answer

The wavelength of Pulse 1 becomes longer when it reaches point $$A$$

Answer

The velocity of any pulse along the rope is independent of its frequency and wavelength

Waves

Choose the correct answer: