1. Analyze the physical setup and derive the wave velocity

Let's set up a coordinate system for the rope. Let the bottom end of the rope (point A, where block M is attached) be at position y=0, and the top end (point O, the fixed support) be at y=L, where L is the length of the rope. The rope has a uniform mass per unit length, μ.

The tension T in the rope at a height y from the bottom end is the sum of the weight of the block M and the weight of the portion of the rope below y.

Weight of the block = Mg Mass of the rope below y = μy Weight of the rope below y = (μy)g

So, the tension at position y is: $T(y) = Mg + μyg = g(M + μy)$

The velocity v of a transverse wave on a rope is given by the formula $v = \sqrt{T/μ}$. Substituting the expression for T(y): $v(y) = \sqrt{\frac{g(M + μy)}{μ}} = \sqrt{g\left(\frac{M}{μ} + y\right)}$ This equation shows that the wave velocity is not constant; it depends on the position y. The velocity is greater at the top of the rope (larger y) and smaller at the bottom.

2. Evaluate each option

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

The time dt it takes for a pulse to travel a small distance dy is given by dt = dy/v(y). To find the total travel time, we integrate this expression over the length of the rope, L.

Time for Pulse 1 to travel from O (y=L) to A (y=0), $T_{OA}$: Since speed v(y) is always positive, the time taken is the integral of ds/v over the path length. $T_{OA} = \int_{0}^{L} \frac{dy}{v(y)} = \int_{0}^{L} \frac{dy}{\sqrt{g\left(\frac{M}{μ} + y\right)}}$

Time for Pulse 2 to travel from A (y=0) to O (y=L), $T_{AO}$: $T_{AO} = \int_{0}^{L} \frac{dy}{v(y)} = \int_{0}^{L} \frac{dy}{\sqrt{g\left(\frac{M}{μ} + y\right)}}$

The mathematical expressions for $T_{OA}$ and $T_{AO}$ are identical because the speed of the wave at any point y is the same regardless of the direction of propagation. Therefore, $T_{AO} = T_{OA}$.

Conclusion: Option A is correct.

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

The midpoint of the rope is at y = L/2. The speed of any pulse at this point is: $v(L/2) = \sqrt{g\left(\frac{M}{μ} + \frac{L}{2}\right)}$ Pulse 1 (from O to A) travels downwards. Its velocity at the midpoint is a vector pointing in the -y direction: $\vec{v}_1 = -v(L/2) \hat{j}$. Pulse 2 (from A to O) travels upwards. Its velocity at the midpoint is a vector pointing in the +y direction: $\vec{v}_2 = +v(L/2) \hat{j}$.

Since velocity is a vector quantity, $\vec{v}_1 \neq \vec{v}_2$ because they have opposite directions. While their speeds are equal, their velocities are not.

Conclusion: Option B is incorrect.

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

When a wave propagates through a medium where the properties change, its frequency f remains constant, determined by the source. The relationship between velocity, frequency, and wavelength is v = fλ. This means the wavelength λ is directly proportional to the velocity v (λ = v/f).

Pulse 1 starts at O (y=L) and travels to A (y=0). Velocity at O: $v_O = v(L) = \sqrt{g(M/μ + L)}$ Velocity at A: $v_A = v(0) = \sqrt{g(M/μ)}$

Since L > 0, it is clear that $v_O > v_A$. The velocity of the pulse decreases as it travels from top to bottom.

Because λ ∝ v, the wavelength also decreases. The initial wavelength at O is $λ_0 = v_O/f$, and the final wavelength at A is $λ_A = v_A/f$. Since $v_O > v_A$, it follows that $λ_0 > λ_A$. The wavelength of Pulse 1 becomes shorter, not longer, when it reaches point A.

Conclusion: Option C is incorrect.

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

The velocity of the pulse at any point y is given by $v(y) = \sqrt{g(M/μ + y)}$. This expression depends on the properties of the rope (μ), the attached mass (M), acceleration due to gravity (g), and the position (y). It does not depend on the wave's characteristics like frequency f or wavelength λ. This is a feature of wave propagation in a non-dispersive medium, which an ideal rope is considered to be.

Conclusion: Option D is correct.