Transverse Wave Pulse Propagation along Hanging Rope

To determine the correct options, let us analyze the mechanics of transverse wave propagation along a hanging uniform rope carrying a suspended block $M$.

1. Tension and Wave Speed along the Rope

Let $y$ be the vertical distance measured from the lower end $A$ of the rope ($y = 0$ at point $A$ and $y = L$ at the top support $O$).

Let $\mu$ be the uniform linear mass density (mass per unit length) of the rope, and $g$ be the acceleration due to gravity. The tension $T(y)$ in the rope at a height $y$ from the bottom end $A$ is given by the sum of the weight of the block $M$ and the weight of the segment of the rope below height $y$: $T(y) = Mg + \mu y g$

The velocity $v(y)$ of a transverse wave pulse at any point $y$ on the rope is determined strictly by the local tension $T(y)$ and the linear mass density $\mu$: $v(y) = \sqrt{\frac{T(y)}{\mu}} = \sqrt{\frac{Mg + \mu y g}{\mu}} = \sqrt{\frac{Mg}{\mu} + gy}$

---

2. Analysis of Options

• Option (A): The time taken by Pulse 1 to travel from $O$ to $A$ ($y = L$ to $y = 0$) is: $T_{OA} = \int_0^L \frac{dy}{v(y)}$ The time taken by Pulse 2 to travel from $A$ to $O$ ($y = 0$ to $y = L$) is: $T_{AO} = \int_0^L \frac{dy}{v(y)}$ Since the local speed $v(y)$ at any point $y$ is independent of the direction of propagation, the two integrals are identical: $T_{AO} = T_{OA}$ Thus, Option (A) is correct.

• Option (B): At the midpoint of the rope ($y = L/2$), the speed $v(L/2)$ of both pulses is identical. However, velocity is a vector quantity. Pulse 1 moves vertically downward: $\vec{v}_1 = -v\left(\frac{L}{2}\right)\hat{j}$ Pulse 2 moves vertically upward: $\vec{v}_2 = +v\left(\frac{L}{2}\right)\hat{j}$ Since $\vec{v}_1 = -\vec{v}_2 \neq \vec{v}_2$, the velocities of the two pulses are opposite in direction and therefore not equal. Thus, Option (B) is incorrect.

• Option (C): The frequency $f$ of a wave pulse is determined solely by its source and remains constant as the pulse travels through the medium. The relationship between velocity $v$, frequency $f$, and wavelength $\lambda$ is: $v = f \lambda \implies \lambda(y) = \frac{v(y)}{f}$ As Pulse 1 propagates from $O$ ($y = L$) to $A$ ($y = 0$), the height $y$ decreases, which causes the tension $T(y)$ and consequently the speed $v(y)$ to decrease. Since $v(y)$ decreases and $f$ remains constant, the wavelength $\lambda(y)$ decreases as Pulse 1 approaches point $A$. Thus, Option (C) is incorrect.

• Option (D): The speed of a transverse wave in a flexible string/rope is given by $v = \sqrt{T/\mu}$. This speed is determined entirely by the physical properties of the medium (tension $T$ and mass density $\mu$) and is completely independent of the frequency $f$ or wavelength $\lambda$ of the wave pulse. Thus, Option (D) is correct.

---

Conclusion

The correct statements are A and D.