A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at `x=2` cm is found at its mean position and it is moving in positive y-direction at `t=1` s. the amplitude of the wave, the wavelength and the angular frequency of the wave are `0.1m,pi//4m` and `4pi rad//s`, respectively.
The speed of particle at `x=2 m` and `t=1 s` is

Options

A. `0.2pi m//s`
B. `0.6pi m//s`
C. `0.4pi m//s`
D. 0

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given parameters
We have the following parameters from the problem statement:
- Amplitude (A) = 0.1 m
- Wavelength (λ) = π/4 m
- Angular frequency (ω) = 4π rad/s
- The position of the particle (x) = 2 cm = 0.02 m (converted to meters)
- Time (t) = 1 s

### Step 2: Calculate the wave number (k)
The wave number (k) can be calculated using the formula:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of λ:
\[ k = \frac{2\pi}{\frac{\pi}{4}} = 8 \, \text{m}^{-1} \]

### Step 3: Calculate the frequency (f)
The frequency (f) can be calculated using the relationship between angular frequency and frequency:
\[ \omega = 2\pi f \]
Rearranging gives:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of ω:
\[ f = \frac{4\pi}{2\pi} = 2 \, \text{Hz} \]

### Step 4: Calculate the speed of the wave (v)
The speed of the wave can be calculated using the formula:
\[ v = f \cdot \lambda \]
Substituting the values of f and λ:
\[ v = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \, \text{m/s} \]

### Step 5: Determine the velocity of the particle at x = 2 m and t = 1 s
Since the particle at x = 2 cm (0.02 m) is at its mean position, we can conclude that the particle at x = 2 m is also at its mean position (as it is an integral multiple of the wavelength). At the mean position, the particle has maximum velocity, which is given by:
\[ v_{\text{max}} = A \cdot \omega \]
Substituting the values of A and ω:
\[ v_{\text{max}} = 0.1 \cdot 4\pi = 0.4\pi \, \text{m/s} \]

### Final Answer
The speed of the particle at x = 2 m and t = 1 s is:
\[ v = 0.4\pi \, \text{m/s} \]

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