A wire having length \(\mathrm{L}\) is kept under tension between \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\). In one experiment, the equation of the wave and energy is given by $\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}\( and \)\mathrm{E}_{1}$ respectively. In another experiment, it is \(\mathrm{y}_{2}=\mathrm{A} \sin\) \(\\{(2 \pi \mathrm{x}) / \mathrm{L}\\} \sin 2 \omega \mathrm{t}\) and \(\mathrm{E}_{2}\). Then........ (A) \(E_{2}=E_{1}\) (B) \(E_{2}=2 \mathrm{E}_{1}\) (C) \(\mathrm{E}_{2}=4 \mathrm{E}_{1}\) (D) \(E_{2}=16 \mathrm{E}_{1}\)

We are given the wave equation for the first experiment as: \(y_1 = A \sin(\frac{\pi x}{L}) \sin(\omega t)\) To find the energy expression for this wave, we know that the wave energy can be expressed as: \(E = \frac{1}{2} \mu \int_0^L (\frac{\partial y}{\partial t})^2 dx\) First, we need to find the time derivative of the wave equation: \(\frac{\partial y_1}{\partial t} = A \sin(\frac{\pi x}{L}) \cos(\omega t) \cdot \omega\) Now, we will square the time derivative: \((\frac{\partial y_1}{\partial t})^2 = \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t)\) Now we can find the energy expression for the first experiment by plugging this value into the energy expression formula: \(E_1 = \frac{1}{2} \mu \int_0^L \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t) dx\)

We are given the wave equation for the second experiment as: \(y_2 = A \sin(\frac{2 \pi x}{L}) \sin(2\omega t)\) Now, similar to the first wave, we will find the time derivative of the wave equation: \(\frac{\partial y_2}{\partial t} = A \sin(\frac{2 \pi x}{L}) \cos(2\omega t) \cdot 2\omega\) Then we square the time derivative: \((\frac{\partial y_2}{\partial t})^2 = 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t)\) Now we can find the energy expression for the second experiment: \(E_2 = \frac{1}{2} \mu \int_0^L 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t) dx\)

Now we have the energy expressions for both the first and second experiments: \(E_1 = \frac{1}{2} \mu \int_0^L \omega^2 A^2 \sin^2(\frac{\pi x}{L})\cos^2(\omega t) dx\) \(E_2 = \frac{1}{2} \mu \int_0^L 4\omega^2 A^2 \sin^2(\frac{2\pi x}{L})\cos^2(2\omega t) dx\) We can see that the only difference between these two expressions are the coefficients of \(\omega^2 A^2\) in the integral: First integral has \(\omega^2 A^2\), while the second integral has \(4\omega^2 A^2\). Therefore, the energy expressions are related to each other as: \(E_2 = 4E_1\) Thus, the correct answer is: (C) \(E_2=4E_1\)