Find the time taken by a transverse wave to travel the full length fo a uniform rope of mass 0.1 kg and length 2.45 m hangs from the ceiling

Options

A. 1 s
B. 0.5 s
C. 2 s
D. 1.5 s

Explanation

To find the time taken by a transverse wave to travel the full length of a uniform rope, we can follow these steps:

### Step 1: Determine the mass per unit length (μ) of the rope
The mass per unit length (μ) can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
where \(m\) is the mass of the rope and \(L\) is its length.

Given:
- Mass, \(m = 0.1 \, \text{kg}\)
- Length, \(L = 2.45 \, \text{m}\)

Calculating:
\[ \mu = \frac{0.1 \, \text{kg}}{2.45 \, \text{m}} \approx 0.0408163 \, \text{kg/m} \]

### Step 2: Understand the relationship between tension and mass
The tension \(T\) in the rope at a distance \(x\) from the bottom can be expressed as:
\[ T = \mu \cdot x \cdot g \]
where \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).

### Step 3: Determine the wave velocity (v)
The wave velocity \(v\) at a distance \(x\) is given by:
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu \cdot x \cdot g}{\mu}} = \sqrt{x \cdot g} \]

### Step 4: Set up the time differential (dt)
The time taken \(dt\) to travel a small distance \(dx\) is:
\[ dt = \frac{dx}{v} = \frac{dx}{\sqrt{x \cdot g}} \]

### Step 5: Integrate to find total time (t)
To find the total time \(t\) taken to travel the entire length \(L\), we integrate:
\[ t = \int_0^L \frac{dx}{\sqrt{x \cdot g}} \]

This can be simplified as:
\[ t = \frac{1}{\sqrt{g}} \int_0^L \frac{dx}{\sqrt{x}} \]

The integral \(\int \frac{dx}{\sqrt{x}}\) evaluates to \(2\sqrt{x}\), so we have:
\[ t = \frac{1}{\sqrt{g}} \left[ 2\sqrt{x} \right]_0^L = \frac{2}{\sqrt{g}} \sqrt{L} \]

### Step 6: Substitute values to find the time
Substituting \(L = 2.45 \, \text{m}\) and \(g = 9.8 \, \text{m/s}^2\):
\[ t = 2 \sqrt{\frac{2.45}{9.8}} = 2 \sqrt{0.25} = 2 \cdot 0.5 = 1 \, \text{s} \]

### Final Answer
The time taken by the transverse wave to travel the full length of the rope is approximately **1 second**.

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