A string of length 3L is fixed at both ends it resonates with a tunning fork in third harmonic with amplitude at antinode equal to `A_(0)` at time t=0 a string element at position of antinode is at half its positive amplitude and moving towards mean position displacement of a string element at `L//2` is given by

Options

A. `A_(0)/2sin(omegat+(11pi)/6)`
B. `sqrt(3A_(0))/2sin(omegat+(5pi)/6)`
C. `A_(0)sin(omegat+(5pi)/6)`
D. `A_(0)/2sin(omegat+(5pi)/6)`

Explanation

To solve the problem step by step, we need to analyze the situation described:

### Step 1: Understand the Harmonics
The string is fixed at both ends and resonates in the third harmonic. The length of the string is \(3L\). In the third harmonic, the number of antinodes is 3, and the wavelength \(\lambda\) can be determined as follows:
\[ \text{Length of the string} = \text{Number of half wavelengths} \times \frac{\lambda}{2} \]
For the third harmonic, we have:
\[ 3L = 3 \times \frac{\lambda}{2} \implies \lambda = 2L \]

### Step 2: Determine the Wave Equation
The wave equation for a standing wave on the string can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t + \phi) \]
where:
- \(A\) is the amplitude at the antinode (given as \(A_0\)),
- \(k\) is the wave number, \(k = \frac{2\pi}{\lambda} = \frac{2\pi}{2L} = \frac{\pi}{L}\),
- \(\omega\) is the angular frequency.

### Step 3: Identify the Position of Interest
We are interested in the displacement of a string element at position \(L/2\). Substituting \(x = L/2\) into the wave equation gives:
\[ y\left(\frac{L}{2}, t\right) = A_0 \sin\left(\frac{\pi}{L} \cdot \frac{L}{2}\right) \cos(\omega t + \phi) \]
This simplifies to:
\[ y\left(\frac{L}{2}, t\right) = A_0 \sin\left(\frac{\pi}{2}\right) \cos(\omega t + \phi) = A_0 \cos(\omega t + \phi) \]

### Step 4: Determine the Phase \(\phi\)
At time \(t = 0\), the string element at the antinode is at half its positive amplitude and moving towards the mean position. This means:
- The displacement at \(t = 0\) is \(y\left(\frac{L}{2}, 0\right) = \frac{A_0}{2}\).
- The velocity is directed towards the mean position, indicating that the particle is moving downwards.

Using the equation:
\[ \frac{A_0}{2} = A_0 \cos(\phi) \implies \cos(\phi) = \frac{1}{2} \implies \phi = \frac{\pi}{3} \text{ or } \phi = \frac{5\pi}{3} \]
Since the particle is moving towards the mean position, we take \(\phi = \frac{5\pi}{3}\).

### Step 5: Final Displacement Equation
Substituting \(\phi\) back into the displacement equation:
\[ y\left(\frac{L}{2}, t\right) = A_0 \cos(\omega t + \frac{5\pi}{3}) \]

### Conclusion
The displacement of a string element at \(L/2\) is given by:
\[ y\left(\frac{L}{2}, t\right) = A_0 \cos(\omega t + \frac{5\pi}{3}) \]