In (Q. 10.) the relative deformation amplitude of medium is

Options

A. `0.02pi`
B. `0.08pi`
C. `0.06pi`
D. None of these

Explanation

To solve the problem of finding the relative deformation amplitude of the medium given the wave equation, we can follow these steps:

### Step 1: Understand the Wave Equation
The wave equation provided is:
\[ y = 4 \sin\left( \frac{2\pi}{0.02} t - \frac{2\pi}{100} x \right) \]
Here, \( y \) is the displacement of the medium, \( t \) is time, and \( x \) is the position along the string.

### Step 2: Identify the Parameters
From the wave equation:
- The amplitude \( A \) is 4 cm.
- The angular frequency \( \omega \) can be derived from the term \( \frac{2\pi}{0.02} \).
- The wave number \( k \) can be derived from the term \( \frac{2\pi}{100} \).

### Step 3: Calculate the Wave Number and Angular Frequency
- The wave number \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{100} \]
Thus, the wavelength \( \lambda \) is:
\[ \lambda = 100 \, \text{cm} \]

- The angular frequency \( \omega \) is:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{0.02} \]
where \( T \) is the period of the wave.

### Step 4: Find the Partial Derivative of y with respect to x
To find the relative deformation amplitude, we need to calculate the partial derivative \( \frac{\partial y}{\partial x} \):
\[ \frac{\partial y}{\partial x} = 4 \cdot \left(-\frac{2\pi}{100}\right) \cos\left( \frac{2\pi}{0.02} t - \frac{2\pi}{100} x \right) \]

### Step 5: Determine the Maximum Value of the Derivative
The maximum value of \( \frac{\partial y}{\partial x} \) occurs when the cosine term is either 1 or -1:
\[ \text{Maximum} \left( \frac{\partial y}{\partial x} \right) = 4 \cdot \frac{2\pi}{100} = \frac{8\pi}{100} \]

### Step 6: Calculate the Relative Deformation Amplitude
Now, we can calculate the relative deformation amplitude:
\[ \text{Relative Deformation Amplitude} = \frac{8\pi}{100} \]

### Step 7: Numerical Calculation
Using \( \pi \approx 3.14 \):
\[ \frac{8 \cdot 3.14}{100} = \frac{25.12}{100} = 0.2512 \]

### Final Answer
Thus, the relative deformation amplitude of the medium is approximately:
\[ 0.2512 \, \text{(dimensionless)} \]