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: Identify the given parameters
The wave equation is given as:
\[ y = A \sin\left(\frac{2\pi}{\lambda} x - \omega t\right) \]
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: Calculate the length of the wave segment passing through the origin
To find the length \( L \) of the wave segment passing through the origin during the time interval from \( t = 0 \) to \( t = \frac{\pi}{2\omega} \), we note that at \( t = 0 \):
\[ y = A \sin\left(\frac{2\pi}{\lambda} x\right) \]
And 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 at \( t = 0 \) and \( t = \frac{\pi}{2\omega} \) indicates that the wave travels a distance of \( \frac{\lambda}{4} \) during this time.

### Step 3: Calculate the total mechanical energy
The total mechanical energy \( E \) in a segment of the wave can be expressed as:
\[ E = \frac{1}{2} \rho L A^2 \omega^2 \]
Where \( L = \frac{\lambda}{4} \). Substituting \( L \):
\[ E = \frac{1}{2} \rho \left(\frac{\lambda}{4}\right) A^2 \omega^2 \]

### Step 4: Substitute the values
Now substituting the values into the equation:
\[ E = \frac{1}{2} \left(3 \times 10^{-2}\right) \left(\frac{0.16}{4}\right) \left(1 \times 10^{-3}\right)^2 (100)^2 \]

### Step 5: Simplify the expression
Calculating each part:
- \( \frac{0.16}{4} = 0.04 \)
- \( (1 \times 10^{-3})^2 = 1 \times 10^{-6} \)
- \( 100^2 = 10000 \)

Now substituting these values:
\[ E = \frac{1}{2} \times (3 \times 10^{-2}) \times (0.04) \times (1 \times 10^{-6}) \times (10000) \]

Calculating:
\[ E = \frac{1}{2} \times 3 \times 10^{-2} \times 0.04 \times 10^{-2} \]
\[ = \frac{1}{2} \times 3 \times 0.04 \times 10^{-4} \]
\[ = 0.06 \times 10^{-4} \, \text{J} = 6 \times 10^{-6} \, \text{J} = 6 \, \mu J \]

### Final Answer
The total energy passing through the origin from \( t = 0 \) to \( t = \frac{\pi}{2\omega} \) is:
\[ \boxed{6 \, \mu J} \]