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

Q: Why doesn't the frequency change as the wave moves up?

Frequency is determined entirely by the source that creates the wave. Think of it like a metronome; once it sets the rhythm, the wave must follow that rhythm regardless of the medium it enters.

Q: Why does the tension change? I thought tension in a string is constant.

Tension is only constant in a *massless* string. Here, the rope has a mass of $6\text{ kg}$. Every segment of the rope must support the weight of all the rope below it, plus the block. So, tension increases as you go up.

Q: Does the linear mass density $\mu$ change?

No, the problem states it is a "uniform thin rope", which means its mass is distributed evenly along its length. Therefore, $\mu$ is constant throughout.

Q: Why did we ignore the length of the rope in our final calculation?

The length of the rope determines the total mass ($6\text{ kg}$), which we used to find the top tension. The $12\text{ m}$ length itself isn't directly needed in the ratio $\sqrt{T_2/T_1}$ because we already have the total mass.

Q: What would happen if the block was much heavier than the rope?

If the block's mass was much greater than the rope's mass, the tension $T_1$ and $T_2$ would be almost equal. The wave speed would be nearly constant, and the wavelength would barely change!

Question

A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope?

Options:

(A) 3

(B) 6

(C) 12

(D) 9

Sigma Percentile · Accepted Answer

### The Anatomy of a Massive Rope

Imagine you are standing in a laboratory, looking at a thick, heavy rope hanging from the ceiling.

This is not your typical idealized physics problem where the string is magically massless. This rope has a substantial mass of $6	ext{ kg}$ and stretches down for $12	ext{ m}$.

At the very bottom of this rope, a $2	ext{ kg}$ block is firmly attached, pulling it taut.

When a wave is generated at the bottom, it begins its journey upwards. But because the rope has mass, the environment the wave travels through is constantly changing.

### The Speed of a Wave

To understand what happens to the wave, we need to look at the fundamental equation for the speed of a transverse wave on a stretched string:

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

Here, $T$ is the tension in the string, and $\mu$ is the linear mass density.

As the wave travels upwards, the frequency $f$ remains absolutely constant. Why? Because frequency is a property of the **source** that generated the wave. Once the wave is born, its frequency is locked in.

However, the wave speed $v$ is determined by the medium. Since $v = f\lambda$, and $f$ is constant, the wavelength $\lambda$ is directly proportional to the wave speed.

Therefore, the wavelength is directly proportional to the square root of the tension:

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

### Calculating the Tension Gradient

Because the rope has mass, the tension is not uniform. It increases as you move higher up the rope.

Let's analyze the tension at the very bottom, $T_1$. At this point, the rope only needs to support the weight of the $2	ext{ kg}$ block hanging below it.

$$T_1 = m_{	ext{block}}g = 2g$$

Now, let's look at the very top of the rope, where it attaches to the ceiling. The support here must hold up the entire weight of the rope plus the block.

$$T_2 = (m_{	ext{rope}} + m_{	ext{block}})g = (6 + 2)g = 8g$$

The tension at the top is four times greater than the tension at the bottom!

### The Wavelength Transformation

Since we know that the wavelength is proportional to the square root of the tension, we can set up a simple ratio to find the new wavelength at the top, $\lambda_2$.

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$$

Substituting the tension values we just calculated:

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2$$

This elegant result tells us that the wavelength at the top of the rope is exactly twice the wavelength at the bottom.

### The Final Verdict

The problem states that the initial wavelength at the bottom, $\lambda_1$, is $6	ext{ cm}$.

Using our ratio, we can easily find the final wavelength:

$$\lambda_2 = 2 	imes 6	ext{ cm} = 12	ext{ cm}$$

As the wave travels up the massive rope, it enters regions of higher and higher tension. This increased tension causes the wave to speed up, and because the frequency must remain constant, the wave stretches out, doubling its wavelength by the time it reaches the ceiling.

Physics is beautifully consistent!

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Visualized Solution

$S y s t e m S e t u p$

$W a v e S p ee d an d T e n s i o n$

$T e n s i o na tt h e B o tt o m$

$T e n s i o na tt h e T o p$

$R a t i oo f W a v e l e n g t h s$

$F ina l W a v e l e n g t h$

$F oo df or T h o ug h t$

The Sigma Insight: Wave Equation and Wave Speed

The Anatomy of a Massive Rope

The Speed of a Wave

Calculating the Tension Gradient

The Wavelength Transformation

The Final Verdict

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Visualized Solution

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