A string of length 'L' is fixed at both ends . It is vibrating in its ` 3rd` overtone with maximum amplitude 'a'. The amplitude at a distance `L//3` from one end is

Options

A. a
B. 0
C. `(sqrt3a)/ 2`
D. `a/2`

Answer

To solve the problem of finding the amplitude at a distance \( \frac{L}{3} \) from one end of a string vibrating in its third overtone, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wavelength**:
The third overtone corresponds to the 4th harmonic (since the overtone number is one less than the harmonic number). The relationship between the wavelength \( \lambda \) and the length \( L \) of the string is given by:
\[ L = 4 \cdot \frac{\lambda}{2} \quad \Rightarrow \quad L = 2\lambda \]
Therefore, we can express the wavelength as:
\[ \lambda = \frac{L}{2} \]

2. **Amplitude Function**:
The amplitude \( A(x) \) of a vibrating string can be expressed as:
\[ A(x) = a \sin\left(\frac{2\pi}{\lambda} x\right) \]
where \( a \) is the maximum amplitude.

3. **Substituting for \( \lambda \)**:
Substitute \( \lambda = \frac{L}{2} \) into the amplitude function:
\[ A(x) = a \sin\left(\frac{2\pi}{\frac{L}{2}} x\right) = a \sin\left(\frac{4\pi}{L} x\right) \]

4. **Calculate Amplitude at \( x = \frac{L}{3} \)**:
Now, we need to find the amplitude at a distance \( x = \frac{L}{3} \):
\[ A\left(\frac{L}{3}\right) = a \sin\left(\frac{4\pi}{L} \cdot \frac{L}{3}\right) = a \sin\left(\frac{4\pi}{3}\right) \]

5. **Evaluate \( \sin\left(\frac{4\pi}{3}\right) \)**:
The value of \( \sin\left(\frac{4\pi}{3}\right) \) can be calculated:
\[ \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
Thus, the amplitude at \( x = \frac{L}{3} \) is:
\[ A\left(\frac{L}{3}\right) = a \left(-\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{2} a \]
Since amplitude cannot be negative, we consider the magnitude:
\[ A\left(\frac{L}{3}\right) = \frac{\sqrt{3}}{2} a \]

### Final Answer:
The amplitude at a distance \( \frac{L}{3} \) from one end is:
\[ \frac{\sqrt{3}}{2} a \]