The equation of a trevelling wave In a uniform string of mass per unit length unit length `mu` is given as `y=Asin(omega-kx)`.Find the total energy transferred through the origin in time interval from t = 0 to `t=pi//12omega.`(You can use the formula in instantaneous power if you know)

Answer

To solve the problem, we need to find the total energy transferred through the origin in the time interval from \( t = 0 \) to \( t = \frac{\pi}{12\omega} \) for the traveling wave described by the equation \( y = A \sin(\omega t - kx) \).

### Step-by-Step Solution:

1. **Identify the Wave Equation**:
The wave equation is given as:
\[ y = A \sin(\omega t - kx) \]
Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

2. **Calculate the Velocity of the Wave**:
The velocity \( v \) of the wave can be derived from the wave equation. The velocity of a wave on a string is given by:
\[ v = \frac{\partial y}{\partial t} = A\omega \cos(\omega t - kx) \]

3. **Determine the Instantaneous Power**:
The instantaneous power \( P \) transmitted by the wave is given by:
\[ P = \frac{1}{2} \mu v \omega^2 \]
Substituting \( v \) into the power equation:
\[ P = \frac{1}{2} \mu (A \omega \cos(\omega t - kx)) \omega^2 = \frac{1}{2} \mu A \omega^3 \cos(\omega t - kx) \]

4. **Calculate Total Energy Transferred**:
The total energy \( E \) transferred through the origin from \( t = 0 \) to \( t = \frac{\pi}{12\omega} \) can be calculated by integrating the power over time:
\[ E = \int_0^{\frac{\pi}{12\omega}} P \, dt = \int_0^{\frac{\pi}{12\omega}} \frac{1}{2} \mu A \omega^3 \cos(\omega t - kx) \, dt \]

5. **Evaluate the Integral**:
The integral can be evaluated as follows:
\[ E = \frac{1}{2} \mu A \omega^3 \int_0^{\frac{\pi}{12\omega}} \cos(\omega t - kx) \, dt \]
The integral of \( \cos \) can be computed:
\[ \int \cos(\omega t - kx) \, dt = \frac{1}{\omega} \sin(\omega t - kx) \]
Thus,
\[ E = \frac{1}{2} \mu A \omega^3 \left[ \frac{1}{\omega} \sin(\omega t - kx) \right]_0^{\frac{\pi}{12\omega}} \]

6. **Substituting the Limits**:
Evaluating the limits:
\[ E = \frac{1}{2} \mu A \omega^3 \left( \frac{1}{\omega} \left( \sin\left(\frac{\pi}{12} - kx\right) - \sin(-kx) \right) \right) \]
Simplifying gives:
\[ E = \frac{1}{2} \mu A \omega^2 \left( \sin\left(\frac{\pi}{12} - kx\right) + \sin(kx) \right) \]

### Final Answer:
The total energy transferred through the origin in the time interval from \( t = 0 \) to \( t = \frac{\pi}{12\omega} \) is:
\[ E = \frac{1}{2} \mu A \omega^2 \left( \sin\left(\frac{\pi}{12} - kx\right) + \sin(kx) \right) \]