A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The

Q: Why is the travel time the same in both directions if one pulse is going down (with gravity) and the other is going up (against gravity)?

The wave speed $v = \sqrt{T/\mu}$ is determined solely by the tension in the rope and its linear mass density. Gravity does not directly accelerate the wave pulse like a physical mass; instead, gravity establishes the static tension profile $T(x)$ in the rope. Since this tension profile is identical whether the wave is traveling up or down, the speed at any point $x$ is the same for both pulses, resulting in equal travel times.

Q: Doesn't the wave pulse carry energy that changes the tension of the rope as it travels?

In the linear wave approximation (which we use for small-amplitude waves), the energy carried by the wave pulse is extremely small compared to the gravitational potential energy of the hanging block and rope. Therefore, the wave pulse does not disturb the static tension distribution of the rope.

Q: Why does the frequency of the wave pulse remain constant as it travels through regions of varying tension?

Frequency is a fundamental property determined by the source that creates the wave. As a wave transition occurs from one region of a medium to another (or through a continuously varying medium), the boundary conditions require the oscillations on both sides to remain in phase. Thus, the frequency must remain constant throughout the propagation.

Q: Are the velocities at the mid-point really the same if the pulses are traveling in opposite directions?

Strictly speaking, velocity is a vector quantity, and since the two pulses are traveling in opposite directions, their velocity vectors point in opposite directions. However, in standard physics problems of this type, the term 'velocity' is often used interchangeably with 'speed' (magnitude of velocity). Under this standard magnitude interpretation, Option (d) is correct.

Q: What would happen to the travel times if the block $M$ was removed?

If the block $M$ was removed ($M = 0$), the tension would be $T(x) = \mu x g$, and the velocity would be $v(x) = \sqrt{gx}$. The travel time would still be equal in both directions because the velocity profile $v(x)$ would still be symmetric and independent of the direction of travel.

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/are correct?

Options:

(A) The time $T_{AO} = T_{OA}$

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

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

(D) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the mid-point of rope

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a heavy block of mass $M$ hanging from a uniform rope of mass $m$ and length $L$.

This is a classic physics scenario where the medium of propagation—the rope—is itself heavy.

Because the rope has mass, the tension is not uniform throughout its length.

Let's set up a coordinate system where the bottom of the rope (where the block is attached) is $x = 0$, and the top of the rope (attached to the ceiling) is $x = L$.

At any arbitrary height $x$ from the bottom, the tension $T(x)$ must support both the hanging block of mass $M$ and the weight of the rope segment of length $x$ below it.

If the mass per unit length of the rope is $\mu$, the mass of this hanging segment is $\mu x$.

Therefore, the tension as a function of height $x$ is:

$$T(x) = Mg + \mu x g$$

This linear variation of tension is the key to understanding everything that follows.

### The Wave Velocity Profile

The speed of a transverse wave on a stretched string is given by the well-known formula:

$$v = \sqrt{\frac{T}{\mu}}$$

Substituting our expression for $T(x)$ into this formula, we get the velocity profile of the wave along the rope:

$$v(x) = \sqrt{\frac{Mg + \mu x g}{\mu}} = \sqrt{\frac{Mg}{\mu} + gx}$$

Notice something incredibly profound here: the velocity of the wave depends **only** on the position $x$ along the rope.

It is completely independent of the wave's frequency $f$ or its wavelength $\lambda$.

This immediately validates **Option (c)** as correct.

### Comparing Travel Times

Now, let's address the travel times for the two pulses.

Pulse 1 is generated at the top ($O$) and travels downwards to the bottom ($A$).

Pulse 2 is generated at the bottom ($A$) and travels upwards to the top ($O$).

Since the position of the block $M$ is undisturbed, the tension profile $T(x)$—and consequently the velocity profile $v(x)$—remains identical for both pulses at any given point $x$.

For an infinitesimal distance $dx$, the time taken is $dt = \frac{dx}{v(x)}$.

To find the total travel time, we integrate this expression over the entire length of the rope from $0$ to $L$:

$$T_{OA} = \int_0^L \frac{dx}{v(x)}$$

$$T_{AO} = \int_0^L \frac{dx}{v(x)}$$

Since the integrand $v(x)$ and the limits of integration are exactly the same, the travel times must be identical:

$$T_{AO} = T_{OA}$$

Thus, **Option (a)** is correct.

### Wavelength Variation

Let's analyze what happens to the wavelength of Pulse 1 as it travels downwards from $O$ to $A$.

As the pulse moves down, the coordinate $x$ decreases.

According to our tension equation, a decrease in $x$ leads to a decrease in tension $T(x)$, which in turn causes the wave speed $v(x)$ to decrease.

Now, the frequency $f$ of a wave is a characteristic of its source and remains constant as the wave propagates through different regions of the medium.

Using the fundamental wave relation:

$$v = f\lambda \implies \lambda = \frac{v}{f}$$

Since $v$ decreases and $f$ remains constant, the wavelength $\lambda$ must also decrease.

Therefore, the wavelength becomes **shorter** as it reaches point $A$, making **Option (b)** incorrect.

### Velocity at the Mid-Point

Finally, let's look at the mid-point of the rope, where $x = L/2$.

At this point, the tension is uniquely determined as:

$$T\left(\frac{L}{2}\right) = Mg + \mu \frac{L}{2} g$$

Both Pulse 1 and Pulse 2 pass through this exact point.

Since the tension and linear mass density are identical for both pulses at this location, their wave speeds must be exactly the same:

$$v_1\left(\frac{L}{2}\right) = v_2\left(\frac{L}{2}\right) = \sqrt{\frac{Mg}{\mu} + \frac{gL}{2}}$$

If we interpret "velocity" in terms of its magnitude (speed), then the velocities of the two pulses are indeed the same at the mid-point.

This confirms **Option (d)** is correct.

Select Answer:

Visualized Solution

Understanding the Physical Setup

Finding Tension $T (x)$ at any Height $x$

Wave Velocity as a Function of $x$

Analyzing Option (c)

Analyzing Option (a)

Analyzing Option (b)

Analyzing Option (d)

Conclusion

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

The Wave Velocity Profile

Comparing Travel Times

Wavelength Variation

Velocity at the Mid-Point

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