One end of a long rope is tied to a fixed vertical pole. The rope is stretched horizontally with a tension `8 N` . Let us cinsider the length of the rope to be along x-axis. A simple harmonic oscillator at `x=0` generates a transverse wave of frequency `100 Hz` and amplitude `2 cm` along the rope. Mass of a unit length of the rope is `20 g//m`. ignoring the effect of gravity, answer the following quetions.
Tension in the given rope remaining the same, if a simple harmonic oscillator of frequency `200 Hz` is used instead of the earlier oscillator of frequency `100 Hz`

Options

A. Speed of transverse waves in the rope will be doubled, wavelength will not change
B. Speed of transverse waves in the rope will become half, wavelength will become one-fourth
C. Speed of transverse waves in the rope will become four times, wavelength will be doubled
D. Speed of transverse waves in the rope will not change, wavelength will become half

Explanation

To solve the problem step by step, we will analyze the information given and apply the relevant physics concepts.

### Step 1: Determine the wave speed in the rope.

The speed of a wave on a stretched string (or rope) can be calculated using the formula:

\[ v = \sqrt{\frac{T}{\mu}} \]

where:
- \( v \) is the wave speed,
- \( T \) is the tension in the rope,
- \( \mu \) is the mass per unit length of the rope.

Given:
- Tension, \( T = 8 \, \text{N} \)
- Mass per unit length, \( \mu = 20 \, \text{g/m} = 0.02 \, \text{kg/m} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))

Substituting the values into the formula:

\[ v = \sqrt{\frac{8 \, \text{N}}{0.02 \, \text{kg/m}}} \]

Calculating the wave speed:

\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Step 2: Analyze the effect of changing frequency on wavelength.

The wave speed \( v \) is related to frequency \( f \) and wavelength \( \lambda \) by the equation:

\[ v = f \lambda \]

From this, we can express wavelength as:

\[ \lambda = \frac{v}{f} \]

Initially, the frequency \( f_1 = 100 \, \text{Hz} \). The new frequency \( f_2 = 200 \, \text{Hz} \).

### Step 3: Calculate the initial wavelength.

Using the initial frequency:

\[ \lambda_1 = \frac{v}{f_1} = \frac{20 \, \text{m/s}}{100 \, \text{Hz}} = 0.2 \, \text{m} \]

### Step 4: Calculate the new wavelength with the new frequency.

Now, using the new frequency:

\[ \lambda_2 = \frac{v}{f_2} = \frac{20 \, \text{m/s}}{200 \, \text{Hz}} = 0.1 \, \text{m} \]

### Conclusion:

1. The speed of the transverse wave in the rope remains constant at \( 20 \, \text{m/s} \).
2. The wavelength changes from \( 0.2 \, \text{m} \) to \( 0.1 \, \text{m} \), which is half of the initial wavelength.

### Final Answer:
- The speed of the transverse wave does not change.
- The wavelength becomes half.

---