A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2

Q: Why does the frequency of the wave pulse remain constant as it travels up the rope?

Frequency is a characteristic of the wave source (the initial disturbance) and does not depend on the properties of the medium. As the wave travels through regions of varying tension, only its speed and wavelength change to keep the frequency constant.

Q: Why is the tension at the bottom of the rope not zero?

The tension at the bottom is not zero because there is a block of mass $M = 2\text{ kg}$ suspended from it. If there were no block, the tension at the very bottom end would indeed be zero.

Q: What would happen to the wave speed at the bottom if there were no suspended block?

If there were no block ($M = 0$), the tension at the bottom would be zero, which means the wave speed at the bottom would also be zero. It would be impossible to start a wave pulse from the bottom without some initial tension!

Q: Does the length of the rope affect the final wavelength in this problem?

No, the length of the rope does not affect the final wavelength. The ratio of the wavelengths depends only on the ratio of the tensions at the top and bottom, which is determined solely by the masses of the block and the rope.

Q: How does the mass density of the rope affect the wave speed?

For a uniform rope, the mass density $\mu$ is constant. A higher mass density decreases the wave speed for a given tension, but since $\mu$ is constant throughout the rope here, it cancels out when we take the ratio of speeds.

Question

A uniform rope of length $12	ext{ m}$ and mass $6	ext{ kg}$ hangs vertically from a rigid support. A block of mass $2	ext{ kg}$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06	ext{ m}$ is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Waves on Hanging Ropes

Imagine a heavy rope hanging from a high ceiling.

If you tap the bottom of the rope, you will see a wave pulse travel upwards, moving faster and faster as it climbs.

Why does this happen?

This phenomenon is a beautiful demonstration of how **gravity, tension, and wave mechanics** intertwine.

In this problem, we explore how the wavelength of a transverse wave pulse changes as it travels from the bottom to the top of a heavy hanging rope with a block suspended at its lower end.

---

### Analyzing the Setup: Gravity's Role in Tension

Let us first understand the physical forces at play.

In a massless string, the tension is uniform throughout its length.

However, for a **heavy rope**, the tension varies with height because each segment of the rope must support the weight of everything hanging below it.

Let the mass of the suspended block be $M = 2	ext{ kg}$ and the mass of the uniform rope be $m = 6	ext{ kg}$.

At the very bottom of the rope ($x = 0$), the rope only supports the suspended block.

Therefore, the tension at the bottom is:

$$T_{	ext{bottom}} = Mg$$

At the very top of the rope ($x = L$), the rope must support both the suspended block and its own entire weight.

Therefore, the tension at the top is:

$$T_{	ext{top}} = (M + m)g$$

---

### The Physics of Wave Speed and Frequency

The speed of a transverse wave on a stretched string or rope is determined by the tension $T$ and the mass per unit length $\mu$:

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

Since the rope is uniform, $\mu$ is constant throughout its length.

As the wave pulse travels upwards, the tension $T$ increases, which means the wave speed $v$ also increases.

But what happens to the frequency $f$ and wavelength $\lambda$?

Here is a fundamental principle of wave propagation: **the frequency of a wave is determined solely by its source and remains constant as the wave travels through a medium.**

Using the fundamental wave relation:

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

Since $f$ is constant, the wavelength $\lambda$ is directly proportional to the wave speed $v$:

$$\lambda \propto v$$

Combining this with the wave speed formula, we find that the wavelength is directly proportional to the square root of the tension:

$$\lambda \propto \sqrt{T}$$

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### Step-by-Step Mathematical Execution

Now, let us set up the ratio of the wavelengths at the top and bottom of the rope:

$$\frac{\lambda_{	ext{top}}}{\lambda_{	ext{bottom}}} = \sqrt{\frac{T_{	ext{top}}}{T_{	ext{bottom}}}}$$

Substitute the expressions for tension into this ratio:

$$\frac{\lambda_{	ext{top}}}{\lambda_{	ext{bottom}}} = \sqrt{\frac{(M + m)g}{Mg}} = \sqrt{\frac{M + m}{M}}$$

Now, substitute the given values ($M = 2	ext{ kg}$ and $m = 6	ext{ kg}$):

$$\frac{\lambda_{	ext{top}}}{\lambda_{	ext{bottom}}} = \sqrt{\frac{2 + 6}{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$$

This elegant result tells us that the wavelength at the top is exactly **twice** the wavelength at the bottom!

$$\lambda_{	ext{top}} = 2 	imes \lambda_{	ext{bottom}}$$

Given that the initial wavelength at the bottom is $\lambda_{	ext{bottom}} = 0.06	ext{ m}$:

$$\lambda_{	ext{top}} = 2 	imes 0.06	ext{ m} = 0.12	ext{ m}$$

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### Deep Dive: Calculating the Travel Time

A classic and thrilling extension of this problem is to calculate the total time $t$ taken for the pulse to travel from the bottom to the top of the rope.

At any distance $x$ from the bottom, the tension is:

$$T(x) = Mg + \left(\frac{m}{L}ight)x g$$

The wave speed at position $x$ is:

$$v(x) = \sqrt{\frac{T(x)}{\mu}} = \sqrt{\frac{Mg + \mu x g}{\mu}} = \sqrt{\frac{M}{\mu}g + xg}$$

Since $v = \frac{dx}{dt}$, we can find the travel time by integrating:

$$dt = \frac{dx}{v(x)} \implies t = \int_0^L \frac{dx}{\sqrt{\frac{M}{\mu}g + xg}}$$

This integration yields a beautiful result that connects kinematics with wave mechanics, showing how deeply unified physics truly is!

Enter Numerical Value:

Visualized Solution

Visualizing the Physical Setup

The Wave Speed Formula

Frequency Invariance

Connecting Wavelength and Tension

Tension at the Bottom

Tension at the Top

Calculating the Tension Ratio

Calculating the Wavelength Ratio

Finding the Final Wavelength

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

Introduction

Analyzing the Setup

The Physics of Wave Speed and Frequency

Step-by-Step Mathematical Execution

Deep Dive

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