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 a 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/arc correct?

• A. The time $T_{AO}=T_{OA}$
• B. The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope
• C. The wavelength of Pulse 1 becomes longer when it reaches point A
• D. The velocity of any pulse along the rope is independent of its frequency and wavelength

Answer 1

Answer: (A), (B), (D)

The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope
The time $T_{AO}=T_{OA}$
The velocity of any pulse along the rope is independent of its frequency and wavelength

Speed of the wave is directly proportional to the tension $T$ in the string and length $l$.

Speed of the wave $v=\sqrt{\frac{T}{\mu }}$

where $\mu$ is mass of string per unit length.

A. As $T$ and $l$ are same at the similar point

We have average velocity $v=\frac{\int vdt}{dt}=constant$

$T_{AO}=T_{OA}$

B. At mid point, tension for both the waves is same, so speed is same. Hence B is correct.

C. Speed is higher at the point where tension is larger i.e. at O.

Since frequency is constant, wavelength of the wave $\lambda =\frac{v}{\nu }$ $⟹\lambda \propto v$

Thus $\lambda$ is longer at O.

Hence C is incorrect.

D. Velocity depends on tension, not frequency or wavelength. Hence D is also correct.