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

Options

A. a
B. 0
C. `sqrt(3a)/(2)`
D. `a/2`

Answer

To solve the problem, we need to find the amplitude at a distance \( \frac{l}{3} \) from one end of a string that is vibrating in its 3rd overtone.

### Step-by-Step Solution:

1. **Understanding the Overtone**:
The 3rd overtone corresponds to the 4th harmonic (since overtone number \( n \) is equal to \( n + 1 \) harmonic). For a string fixed at both ends, the harmonics are given by the formula:
\[ y(x, t) = A \sin\left(\frac{n \pi x}{l}\right) \cos(\omega t) \]
where \( n \) is the harmonic number, \( A \) is the maximum amplitude, \( x \) is the position along the string, \( l \) is the length of the string, and \( \omega \) is the angular frequency.

2. **Identify Parameters**:
For the 3rd overtone, \( n = 4 \):
\[ y(x, t) = A \sin\left(\frac{4 \pi x}{l}\right) \cos(\omega t) \]

3. **Amplitude at \( x = \frac{l}{3} \)**:
We need to find the amplitude at \( x = \frac{l}{3} \). The amplitude is given by the sine term:
\[ A_{x} = A \sin\left(\frac{4 \pi \left(\frac{l}{3}\right)}{l}\right) \]
Simplifying this:
\[ A_{x} = A \sin\left(\frac{4 \pi}{3}\right) \]

4. **Calculating \( \sin\left(\frac{4 \pi}{3}\right) \)**:
The angle \( \frac{4 \pi}{3} \) is in the third quadrant where sine is negative:
\[ \sin\left(\frac{4 \pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]

5. **Final Amplitude**:
Therefore, the amplitude at a distance \( \frac{l}{3} \) from one end is:
\[ A_{x} = A \left(-\frac{\sqrt{3}}{2}\right) \]
The negative sign indicates the direction of the wave at that point, but the overall amplitude is:
\[ A_{x} = \frac{\sqrt{3}}{2} A \]

### Conclusion:
The amplitude at a distance \( \frac{l}{3} \) from one end is \( \frac{\sqrt{3}}{2} A \).