A string with tension T and mass per unit length `mu` is clamped down at x=0 and at x=L. at t=0, the string is at rest and displaced in the y-direction
`y(x,0)=2"sin"(2pix)/(L)+3"sin"(pix)/(L)`
Q. What is the total energy at t=0?

Options

A. a. `(25Tpi^(2))/(4L)`
B. b. `(25Tpi^(2))/(8L)`
C. c. `(25Tpi^(2))/(2L)`
D. d. `(25Tpi^(2))/(16L)`

Answer

To find the total energy of the string at \( t = 0 \), we can follow these steps:

### Step 1: Understand the Initial Conditions
At \( t = 0 \), the string is at rest and displaced in the y-direction. The displacement is given by:
\[ y(x, 0) = 2 \sin\left(\frac{2\pi x}{L}\right) + 3 \sin\left(\frac{\pi x}{L}\right) \]

### Step 2: Calculate the Potential Energy Density
The potential energy density \( \frac{dU}{dx} \) for a string under tension \( T \) is given by:
\[ \frac{dU}{dx} = \frac{1}{2} T \left( \frac{\partial y}{\partial x} \right)^2 \]

### Step 3: Find the Derivative of Displacement
We need to compute \( \frac{\partial y}{\partial x} \):
\[ \frac{\partial y}{\partial x} = \frac{\partial}{\partial x} \left( 2 \sin\left(\frac{2\pi x}{L}\right) + 3 \sin\left(\frac{\pi x}{L}\right) \right) \]

Using the chain rule:
\[ \frac{\partial y}{\partial x} = 2 \cdot \frac{2\pi}{L} \cos\left(\frac{2\pi x}{L}\right) + 3 \cdot \frac{\pi}{L} \cos\left(\frac{\pi x}{L}\right) \]
\[ = \frac{4\pi}{L} \cos\left(\frac{2\pi x}{L}\right) + \frac{3\pi}{L} \cos\left(\frac{\pi x}{L}\right) \]

### Step 4: Compute the Square of the Derivative
Now, we square the derivative:
\[ \left( \frac{\partial y}{\partial x} \right)^2 = \left( \frac{4\pi}{L} \cos\left(\frac{2\pi x}{L}\right) + \frac{3\pi}{L} \cos\left(\frac{\pi x}{L}\right) \right)^2 \]

### Step 5: Expand the Square
Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ = \left( \frac{4\pi}{L} \cos\left(\frac{2\pi x}{L}\right) \right)^2 + 2 \left( \frac{4\pi}{L} \cos\left(\frac{2\pi x}{L}\right) \right) \left( \frac{3\pi}{L} \cos\left(\frac{\pi x}{L}\right) \right) + \left( \frac{3\pi}{L} \cos\left(\frac{\pi x}{L}\right) \right)^2 \]
\[ = \frac{16\pi^2}{L^2} \cos^2\left(\frac{2\pi x}{L}\right) + \frac{24\pi^2}{L^2} \cos\left(\frac{2\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) + \frac{9\pi^2}{L^2} \cos^2\left(\frac{\pi x}{L}\right) \]

### Step 6: Integrate to Find Total Energy
The total energy \( E \) is given by integrating the potential energy density over the length of the string:
\[ E = \int_0^L \frac{1}{2} T \left( \frac{\partial y}{\partial x} \right)^2 dx \]

Substituting in the expression we derived:
\[ E = \frac{1}{2} T \int_0^L \left( \frac{16\pi^2}{L^2} \cos^2\left(\frac{2\pi x}{L}\right) + \frac{24\pi^2}{L^2} \cos\left(\frac{2\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) + \frac{9\pi^2}{L^2} \cos^2\left(\frac{\pi x}{L}\right) \right) dx \]

### Step 7: Solve the Integrals
Using the identities for the integrals of cosine squared:
\[ \int_0^L \cos^2\left(\frac{n\pi x}{L}\right) dx = \frac{L}{2} \]
and
\[ \int_0^L \cos\left(\frac{2\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) dx = 0 \text{ (orthogonality)} \]
we can compute:
\[ E = \frac{1}{2} T \left( \frac{16\pi^2}{L^2} \cdot \frac{L}{2} + 0 + \frac{9\pi^2}{L^2} \cdot \frac{L}{2} \right) \]
\[ = \frac{1}{2} T \left( \frac{8\pi^2}{L} + \frac{9\pi^2}{2L} \right) \]
\[ = \frac{1}{2} T \cdot \frac{25\pi^2}{4L} \]
\[ = \frac{25T\pi^2}{8L} \]

### Final Answer
The total energy at \( t = 0 \) is:
\[ E = \frac{25T\pi^2}{8L} \]