A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by `y=Asinkxcos(omegat+(pi)/(3))`. All the symbols have their usual meaning. Mass per unit length of the string is `0.5(gm)/(cm)`. Given that `A=1` cm and `omega=100pi(rad)/(sec)` Answer the following 2 questions based on information given (use `mu^(2)=10)`
Q. Starting from `t=0`, energy of vibration is completely potential at time `t`, where t is

Options

A. `(1)/(150)sec`
B. `(1)/(60)sec`
C. `(3)/(100)sec`
D. `(11)/(300)sec`

Answer

To solve the problem, we need to find the time \( t \) at which the energy of vibration of the string is completely potential. This occurs when the displacement of the wave is at its maximum, which corresponds to the cosine term in the wave equation being equal to either \( +1 \) or \( -1 \).

### Step-by-Step Solution:

1. **Understand the Wave Equation**:
The standing wave equation is given by:
\[ y = A \sin(kx) \cos(\omega t + \frac{\pi}{3}) \]
Here, \( A = 1 \, \text{cm} \), \( \omega = 100\pi \, \text{rad/s} \).

2. **Condition for Maximum Displacement**:
For the energy to be completely potential, the displacement \( y \) must be at a maximum. This occurs when:
\[ \cos(\omega t + \frac{\pi}{3}) = \pm 1 \]
This means:
\[ \omega t + \frac{\pi}{3} = n\pi \quad (n \in \mathbb{Z}) \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ \omega t = n\pi - \frac{\pi}{3} \]
Thus:
\[ t = \frac{n\pi - \frac{\pi}{3}}{\omega} \]

4. **Substituting the Value of \( \omega \)**:
Substitute \( \omega = 100\pi \):
\[ t = \frac{n\pi - \frac{\pi}{3}}{100\pi} \]
Simplifying this gives:
\[ t = \frac{n - \frac{1}{3}}{100} \]

5. **Finding Specific Values of \( t \)**:
Now we can find specific values of \( t \) for different integers \( n \):
- For \( n = 0 \):
\[ t = \frac{0 - \frac{1}{3}}{100} = -\frac{1}{300} \, \text{s} \quad (\text{not valid}) \]
- For \( n = 1 \):
\[ t = \frac{1 - \frac{1}{3}}{100} = \frac{2/3}{100} = \frac{2}{300} = \frac{1}{150} \, \text{s} \]
- For \( n = 2 \):
\[ t = \frac{2 - \frac{1}{3}}{100} = \frac{6/3 - 1/3}{100} = \frac{5/3}{100} = \frac{5}{300} = \frac{1}{60} \, \text{s} \]
- For \( n = 3 \):
\[ t = \frac{3 - \frac{1}{3}}{100} = \frac{9/3 - 1/3}{100} = \frac{8/3}{100} = \frac{8}{300} = \frac{2}{75} \, \text{s} \]
- For \( n = 4 \):
\[ t = \frac{4 - \frac{1}{3}}{100} = \frac{12/3 - 1/3}{100} = \frac{11/3}{100} = \frac{11}{300} \, \text{s} \]

### Final Answer:
The times \( t \) at which the energy is completely potential are:
- \( t = \frac{1}{150} \, \text{s} \)
- \( t = \frac{1}{60} \, \text{s} \)
- \( t = \frac{2}{75} \, \text{s} \)
- \( t = \frac{11}{300} \, \text{s} \)