A uniform string is clamped at x=0 and x = L and is vibrating in its fundamental mode. Mass per unit length of the string is `mu`, tension in it is T and the maximum displacement of its midpoint is A. Find the total energy stored in the string. Assume to be small so that changes in tension and length of the string can be ignored

Answer

To find the total energy stored in a uniform string vibrating in its fundamental mode, we can follow these steps:

### Step 1: Understand the setup
The string is clamped at both ends (x = 0 and x = L) and is vibrating in its fundamental mode. This means that the string has nodes at both ends and an antinode at the midpoint.

### Step 2: Determine the wavelength
For a string clamped at both ends, the wavelength (λ) in the fundamental mode is given by:
\[ L = \frac{\lambda}{2} \implies \lambda = 2L \]

### Step 3: Calculate the wave number (k)
The wave number (k) is defined as:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{2L} = \frac{\pi}{L} \]

### Step 4: Write the displacement equation
The displacement of the string at any point x can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t) \]
where A is the amplitude, k is the wave number, and ω is the angular frequency.

### Step 5: Find the angular frequency (ω)
The angular frequency (ω) can be related to the tension (T) and mass per unit length (μ) of the string. The wave speed (v) in the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
The frequency (ν) is related to the wave speed and wavelength:
\[ v = \nu \lambda \implies \nu = \frac{v}{\lambda} = \frac{v}{2L} \]
Thus, the angular frequency is:
\[ \omega = 2\pi \nu = \frac{2\pi}{2L} \sqrt{\frac{T}{\mu}} = \frac{\pi}{L} \sqrt{\frac{T}{\mu}} \]

### Step 6: Calculate the energy in an elemental length (dx)
The energy stored in a small segment of the string (dx) can be expressed as:
\[ dE = \frac{1}{2} dm \cdot A^2 \omega^2 \]
where \(dm = \mu \, dx\). Substituting this in gives:
\[ dE = \frac{1}{2} \mu \, dx \cdot A^2 \omega^2 \]

### Step 7: Substitute ω²
Substituting for ω²:
\[ \omega^2 = \left(\frac{\pi}{L} \sqrt{\frac{T}{\mu}}\right)^2 = \frac{\pi^2 T}{\mu L^2} \]
Thus, we can write:
\[ dE = \frac{1}{2} \mu \, dx \cdot A^2 \cdot \frac{\pi^2 T}{\mu L^2} \]
This simplifies to:
\[ dE = \frac{1}{2} A^2 \frac{\pi^2 T}{L^2} \, dx \]

### Step 8: Integrate to find total energy
To find the total energy (E) in the string, integrate dE from 0 to L:
\[ E = \int_0^L dE = \int_0^L \frac{1}{2} A^2 \frac{\pi^2 T}{L^2} \, dx \]
This gives:
\[ E = \frac{1}{2} A^2 \frac{\pi^2 T}{L^2} \cdot L = \frac{1}{2} A^2 \frac{\pi^2 T}{L} \]

### Final Result
Thus, the total energy stored in the string is:
\[ E = \frac{A^2 \pi^2 T}{4L} \]