A progressive wave on a string having linear mass density `rho` is represented by `y=A sin((2 pi)/(lamda)x-omegat)` where `y` is in mm. Find the total energy (in `mu J`) passing through origin from `t=0` to `t=(pi)/(2 omega)`.
[Take : `rho = 3 xx 10^(-2) kg//m , A = 1mm , omega = 100 rad..sec , lamda = 16 cm`].

Options

A. 6
B. 7
C. 8
D. 9

Explanation

To find the total energy passing through the origin from \( t = 0 \) to \( t = \frac{\pi}{2 \omega} \) for the given progressive wave, we can follow these steps:

### Step 1: Understand the wave equation
The wave is represented by:
\[ y = A \sin\left(\frac{2\pi}{\lambda} x - \omega t\right) \]
where:
- \( A = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- \( \lambda = 16 \text{ cm} = 0.16 \text{ m} \)
- \( \rho = 3 \times 10^{-2} \text{ kg/m} \)
- \( \omega = 100 \text{ rad/s} \)

### Step 2: Determine the length of the wave segment
At \( t = 0 \):
\[ y = A \sin\left(\frac{2\pi}{\lambda} x\right) \]
At \( t = \frac{\pi}{2 \omega} \):
\[ y = A \sin\left(\frac{2\pi}{\lambda} x - \frac{\pi}{2}\right) = A \cos\left(\frac{2\pi}{\lambda} x\right) \]
The wave travels a distance of \( \frac{\lambda}{4} \) during this time.

### Step 3: Calculate the length of the wave segment
The length \( L \) of the wave segment that passes through the origin from \( t = 0 \) to \( t = \frac{\pi}{2 \omega} \) is:
\[ L = \frac{\lambda}{4} = \frac{0.16 \text{ m}}{4} = 0.04 \text{ m} \]

### Step 4: Calculate the total energy
The total mechanical energy \( E \) in a segment of the wave is given by:
\[ E = \frac{1}{2} \cdot \text{mass} \cdot A^2 \cdot \omega^2 \]
The mass of the segment is:
\[ \text{mass} = \rho \cdot L = (3 \times 10^{-2} \text{ kg/m}) \cdot (0.04 \text{ m}) = 1.2 \times 10^{-3} \text{ kg} \]
Now substituting the values into the energy formula:
\[ E = \frac{1}{2} \cdot (1.2 \times 10^{-3}) \cdot (1 \times 10^{-3})^2 \cdot (100)^2 \]
Calculating it step by step:
\[ E = \frac{1}{2} \cdot (1.2 \times 10^{-3}) \cdot (1 \times 10^{-6}) \cdot (10000) \]
\[ E = \frac{1}{2} \cdot (1.2 \times 10^{-3}) \cdot (10^{-2}) \]
\[ E = \frac{1.2 \times 10^{-5}}{2} = 6 \times 10^{-6} \text{ J} = 6 \text{ µJ} \]

### Final Answer
The total energy passing through the origin from \( t = 0 \) to \( t = \frac{\pi}{2 \omega} \) is \( 6 \text{ µJ} \).

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