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.
maximum magnitude of transverse acceleration of any point on the rope will be nearly

Options

A. `7888 m//s^(2)`
B. `8244 m//s^(2)`
C. `9277 m//s^(2)`
D. `3333 m//s^(2)`

Explanation

To find the maximum magnitude of transverse acceleration of any point on the rope, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Tension in the rope, \( T = 8 \, \text{N} \)
- Frequency of the wave, \( f = 100 \, \text{Hz} \)
- Amplitude of the wave, \( A = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Mass per unit length of the rope, \( \mu = 20 \, \text{g/m} = 0.02 \, \text{kg/m} \)

2. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 100 = 200 \pi \, \text{rad/s} \]

3. **Determine Maximum Transverse Acceleration**:
The maximum transverse acceleration \( a_{\text{max}} \) can be calculated using the formula:
\[ a_{\text{max}} = A \omega^2 \]
Substituting the values of \( A \) and \( \omega \):
\[ a_{\text{max}} = 0.02 \times (200 \pi)^2 \]

4. **Calculate \( \omega^2 \)**:
First, calculate \( (200 \pi)^2 \):
\[ (200 \pi)^2 = 40000 \pi^2 \]

5. **Substitute Back to Find \( a_{\text{max}} \)**:
Now substitute \( (200 \pi)^2 \) back into the equation for \( a_{\text{max}} \):
\[ a_{\text{max}} = 0.02 \times 40000 \pi^2 = 800 \pi^2 \]

6. **Approximate the Value**:
Using \( \pi \approx 3.14 \):
\[ a_{\text{max}} \approx 800 \times (3.14)^2 \approx 800 \times 9.8596 \approx 7887.68 \, \text{m/s}^2 \]
Rounding off, we find:
\[ a_{\text{max}} \approx 7888 \, \text{m/s}^2 \]

### Final Answer:
The maximum magnitude of transverse acceleration of any point on the rope will be nearly \( 7888 \, \text{m/s}^2 \).

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