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

Options

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

Explanation

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Understand the relationship between maximum speed, amplitude, and angular frequency
The maximum speed \( v_{max} \) of any point in the string is given by the formula:
\[ v_{max} = A \omega \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 2: Substitute the known values
We know from the problem:
- \( A = 10^{-3} \) m,
- The maximum speed is given as \( \frac{v}{10} \), where \( v = 10 \) m/s. Thus,
\[ v_{max} = \frac{10}{10} = 1 \text{ m/s} \]

Now we can substitute the values into the equation:
\[ 1 = 10^{-3} \omega \]

### Step 3: Solve for angular frequency \( \omega \)
Rearranging the equation gives:
\[ \omega = \frac{1}{10^{-3}} = 10^3 \text{ rad/s} \]

### Step 4: Relate angular frequency to frequency \( f \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:
\[ \omega = 2 \pi f \]
Substituting the value of \( \omega \):
\[ 10^3 = 2 \pi f \]

### Step 5: Solve for frequency \( f \)
Rearranging gives:
\[ f = \frac{10^3}{2 \pi} \text{ Hz} \]

### Step 6: Use the wave speed formula to find wavelength \( \lambda \)
The speed of the wave \( v \) is related to frequency \( f \) and wavelength \( \lambda \) by the equation:
\[ v = f \lambda \]
Rearranging 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 Results
- Frequency \( f = \frac{10^3}{2 \pi} \) Hz
- Wavelength \( \lambda = 2 \pi \times 10^{-2} \) m