A 100 Hz sinusoidal wave is travelling in the positve x - direaction along a string with a linear mass density of `3.5 xx10^(-3) kgm ^(-1)` and a tension of 35 N. At time t =0 ,the point x=0 has zero displacment and the slope of the string is `pi//20`then select the wrong alternative.

Options

A. velcoity of wave is 100 m/s
B. angular velocity is `(200 pi)` rad /s
C. Amplitude of wave is 0.025 m
D. none of the above

Answer

To solve the problem step by step, we will analyze the given information and calculate the required quantities to identify the wrong alternative.

### Given Data:
- Frequency of the wave, \( f = 100 \, \text{Hz} \)
- Linear mass density of the string, \( \mu = 3.5 \times 10^{-3} \, \text{kg/m} \)
- Tension in the string, \( T = 35 \, \text{N} \)
- At time \( t = 0 \), the displacement at \( x = 0 \) is zero, and the slope of the string is \( \frac{\pi}{20} \).

### Step 1: Calculate the Wave Velocity
The wave velocity \( V_w \) can be calculated using the formula:
\[ V_w = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ V_w = \sqrt{\frac{35 \, \text{N}}{3.5 \times 10^{-3} \, \text{kg/m}}} \]
Calculating:
\[ V_w = \sqrt{\frac{35}{0.0035}} = \sqrt{10000} = 100 \, \text{m/s} \]

### Step 2: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the frequency:
\[ \omega = 2\pi \times 100 = 200\pi \, \text{radians/s} \]

### Step 3: Calculate the Amplitude
At \( t = 0 \) and \( x = 0 \), the displacement is zero, which means the particle is at its equilibrium position. The maximum particle velocity \( V_p \) is given by:
\[ V_p = A \omega \]
Where \( A \) is the amplitude. The slope of the string at this point is given by:
\[ \text{slope} = \frac{\partial y}{\partial x} = \frac{\pi}{20} \]
The particle velocity can also be expressed as:
\[ V_p = \text{slope} \times V_w \]
Equating the two expressions for \( V_p \):
\[ A \omega = \left(\frac{\pi}{20}\right) V_w \]
Substituting the values of \( V_w \) and \( \omega \):
\[ A (200\pi) = \left(\frac{\pi}{20}\right) (100) \]
Cancelling \( \pi \) from both sides:
\[ 200A = \frac{100}{20} \]
\[ 200A = 5 \implies A = \frac{5}{200} = 0.025 \, \text{m} \]

### Conclusion
Now we have the following results:
- Wave velocity \( V_w = 100 \, \text{m/s} \)
- Angular frequency \( \omega = 200\pi \, \text{radians/s} \)
- Amplitude \( A = 0.025 \, \text{m} \)

### Identify the Wrong Alternative
Based on the calculations:
- The wave velocity is correct.
- The angular frequency is correct.
- The amplitude is correct.

Thus, the wrong alternative must be option D, as indicated in the video transcript.