In case of a transverses wave in a string

Options

A. if `(lambda)/(2pi) gt A_0` maximum particle speed is more than the wave speed
B. if `(lambda)/(2pi)=A_0`, particle speed is equal to the wave speed
C. if `(lambda)/(2pi)=A_0`, maximum particle speed is less than wave speed
D. if `(lambda)/(2pi)=A_0`, maximum particle speed may be greater than wave speed

Answer

To solve the problem regarding the relationship between maximum particle speed and wave speed in a transverse wave in a string, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Wave Parameters**:
- The wave speed (v) is given by the formula:
\[ v = \frac{\omega}{k} \]
- The maximum particle speed (v_max) is given by:
\[ v_{\text{max}} = \omega A_0 \]
where \( \omega \) is the angular frequency and \( A_0 \) is the amplitude of the wave.

2. **Relating Wavelength and Wave Number**:
- The wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
- Therefore, substituting \( k \) into the wave speed formula gives:
\[ v = \frac{\omega}{\frac{2\pi}{\lambda}} = \frac{\omega \lambda}{2\pi} \]

3. **Setting Up the Conditions**:
- We need to compare the maximum particle speed and wave speed under different conditions of \( A_0 \) and \( \lambda \).
- We will analyze the conditions:
- \( \frac{\lambda}{2\pi} > A_0 \)
- \( \frac{\lambda}{2\pi} = A_0 \)
- \( \frac{\lambda}{2\pi} < A_0 \)

4. **Analyzing Each Condition**:
- **Condition 1**: If \( \frac{\lambda}{2\pi} > A_0 \)
- This implies \( v_{\text{max}} < v \) because:
\[ \frac{\omega A_0}{\frac{\omega \lambda}{2\pi}} < 1 \implies v_{\text{max}} < v \]
- **Condition 2**: If \( \frac{\lambda}{2\pi} = A_0 \)
- This implies \( v_{\text{max}} = v \) because:
\[ \frac{\omega A_0}{\frac{\omega \lambda}{2\pi}} = 1 \implies v_{\text{max}} = v \]
- **Condition 3**: If \( \frac{\lambda}{2\pi} < A_0 \)
- This implies \( v_{\text{max}} > v \) because:
\[ \frac{\omega A_0}{\frac{\omega \lambda}{2\pi}} > 1 \implies v_{\text{max}} > v \]

5. **Conclusion**:
- Based on the analysis:
- If \( \frac{\lambda}{2\pi} > A_0 \), then \( v_{\text{max}} < v \).
- If \( \frac{\lambda}{2\pi} = A_0 \), then \( v_{\text{max}} = v \).
- If \( \frac{\lambda}{2\pi} < A_0 \), then \( v_{\text{max}} > v \).
- Therefore, the correct option is that when \( \frac{\lambda}{2\pi} = A_0 \), the maximum particle speed is equal to the wave speed.