A transverse sinusoidal wave of amplitude A, wavelength `lamda` and frequency f is travelling along a stretch string. The maximum speed of any point on the string is `(v)/(10)`, where, V is the velocity of wave propagation. If `A=10^(-3)m, and V=10ms^(-1)`, then `lamda` and f are given by

Options

A. `lamda=2pixx10^(-2)m`
B. `lamda=10^(-2)m`
C. `f=(10^(3))/(2pi)Hz`
D. `f=10^(4)Hz`

Answer

To solve the problem step by step, we will follow the given information and derive the values of wavelength (λ) and frequency (f) for the transverse sinusoidal wave.

### Given:
- Amplitude (A) = \(10^{-3}\) m
- Velocity of wave propagation (V) = 10 m/s
- Maximum speed of any point on the string = \( \frac{V}{10} = \frac{10}{10} = 1 \) m/s

### Step 1: Find the angular frequency (ω)
The maximum speed of a wave particle is given by the formula:
\[ v_{p_{max}} = A \omega \]
Where:
- \(v_{p_{max}}\) = Maximum speed of the wave particle
- A = Amplitude
- ω = Angular frequency

Substituting the known values into the equation:
\[ 1 = (10^{-3}) \cdot \omega \]
Now, solve for ω:
\[ \omega = \frac{1}{10^{-3}} = 10^3 \text{ rad/s} \]

### Step 2: Find the frequency (f)
The angular frequency (ω) is related to frequency (f) by the equation:
\[ \omega = 2 \pi f \]
Rearranging for f gives:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of ω:
\[ f = \frac{10^3}{2\pi} \]

### Step 3: Find the wavelength (λ)
The relationship between the wave velocity (V), frequency (f), and wavelength (λ) is given by:
\[ V = f \lambda \]
Rearranging for λ gives:
\[ \lambda = \frac{V}{f} \]
Substituting the known values:
\[ \lambda = \frac{10}{\frac{10^3}{2\pi}} = 10 \cdot \frac{2\pi}{10^3} = \frac{2\pi}{10^2} = 2\pi \times 10^{-2} \text{ m} \]

### Final Answers:
- Frequency (f) = \( \frac{10^3}{2\pi} \) Hz
- Wavelength (λ) = \( 2\pi \times 10^{-2} \) m