If the maximum particle velocity is 4 times of the wave velocity then relation between wavelength and amplitude is

Options

A. `2piA`
B. `piA`
C. `(piA)/2`
D. `(piA)/4`

Answer

To solve the problem, we need to establish the relationship between the wavelength (λ) and the amplitude (A) given that the maximum particle velocity (v_max) is 4 times the wave velocity (v).

### Step-by-Step Solution:

1. **Start with the wave equation**:
The wave can be described by the equation:
\[ y = A \sin(\omega t - kx + \phi) \]
where:
- \(y\) is the displacement,
- \(A\) is the amplitude,
- \(\omega\) is the angular frequency,
- \(k\) is the wave number,
- \(x\) is the position,
- \(t\) is the time,
- \(\phi\) is the phase constant.

2. **Differentiate with respect to time**:
To find the particle velocity, we differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = A \omega \cos(\omega t - kx + \phi) \]
The maximum particle velocity (\(v_{max}\)) occurs when \(\cos(\omega t - kx + \phi) = 1\):
\[ v_{max} = A \omega \]

3. **Relate particle velocity to wave velocity**:
According to the problem, the maximum particle velocity is 4 times the wave velocity (\(v\)):
\[ v_{max} = 4v \]
Therefore, we can set up the equation:
\[ A \omega = 4v \]

4. **Express wave velocity**:
The wave velocity \(v\) can be expressed in terms of frequency (\(f\)) and wavelength (\(\lambda\)):
\[ v = f \lambda \]
We can also express the angular frequency \(\omega\) in terms of frequency:
\[ \omega = 2\pi f \]

5. **Substitute \(v\) and \(\omega\) into the equation**:
Substitute \(v\) and \(\omega\) into the equation \(A \omega = 4v\):
\[ A (2\pi f) = 4(f \lambda) \]

6. **Cancel \(f\) from both sides**:
Assuming \(f \neq 0\), we can cancel \(f\):
\[ A (2\pi) = 4\lambda \]

7. **Rearrange to find the relationship**:
Rearranging gives:
\[ \lambda = \frac{2\pi A}{4} = \frac{\pi A}{2} \]

### Final Relationship:
Thus, the relationship between the wavelength and amplitude is:
\[ \lambda = \frac{\pi A}{2} \]