The ends of a stretched wire of length L are fixed at x = 0 and x = L, In one experiment, the displacement of wire is `y_(1) = A sin (pi x//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2)= A sin (2pix//L) sin 2 omegat` and energy is `E_(2)`. Then

Options

A. `E_(2) = E_(1)`
B. `E_(2) = 2E_(1)`
C. `E_(2) = 4E_(1)`
D. `E_(2) = 16 E_(1)`

Explanation

To solve the problem, we need to find the relationship between the energies \( E_1 \) and \( E_2 \) based on the given displacements of the stretched wire in two different experiments.

### Step-by-Step Solution:

1. **Identify the Displacements**:
- For the first experiment, the displacement is given by:
\[ y_1 = A \sin\left(\frac{\pi x}{L}\right) \sin(\omega t) \]
- For the second experiment, the displacement is:
\[ y_2 = A \sin\left(\frac{2\pi x}{L}\right) \sin(2\omega t) \]

2. **Understand the Energy in Waves**:
- The energy per unit volume \( dU \) for a harmonic wave can be expressed as:
\[ dU = \frac{1}{2} \rho A \omega^2 \cos^2(\omega t - kx + \phi) \]
- Here, \( \rho \) is the density, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

3. **Determine the Angular Frequencies**:
- For the first displacement \( y_1 \):
- The angular frequency \( \omega_1 \) is \( \omega \).
- For the second displacement \( y_2 \):
- The angular frequency \( \omega_2 \) is \( 2\omega \).

4. **Calculate the Energies**:
- The energy \( E_1 \) for the first wave is proportional to \( \omega_1^2 \):
\[ E_1 \propto \omega^2 \]
- The energy \( E_2 \) for the second wave is proportional to \( \omega_2^2 \):
\[ E_2 \propto (2\omega)^2 = 4\omega^2 \]

5. **Establish the Relationship**:
- Now, we can relate \( E_1 \) and \( E_2 \):
\[ E_2 \propto 4\omega^2 \]
- Thus, we can express the ratio of energies:
\[ \frac{E_1}{E_2} = \frac{\omega^2}{4\omega^2} = \frac{1}{4} \]
- Therefore, we can conclude:
\[ E_2 = 4E_1 \]

### Final Result:
The relationship between the energies is:
\[ E_2 = 4E_1 \]