A sonometer wire of length `L` is plucked at a distance `L//8` from one end then it vibrates with a minimum frequency `n`. If the same wire plucked at a distance `L//6` from another end the minimum frequency with which it vibrates is

Options

A. `(sqrt(3))/(2)n`
B. `(3)/(2)n`
C. `(3n)/(4)`
D. `(4n)/(3)`

Explanation

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

### Step 1: Understanding the Sonometer Wire
A sonometer wire is a device used to study the vibrations of strings. The frequency of vibration depends on the length of the vibrating segment of the wire. When the wire is plucked at different points, the effective length of the vibrating segment changes, which in turn affects the frequency of vibration.

### Step 2: Frequency Formula
The frequency \( n \) of a vibrating string is given by the formula:
\[ n = \frac{v}{\lambda} \]
where:
- \( v \) is the speed of the wave on the string,
- \( \lambda \) is the wavelength.

### Step 3: First Plucking at \( \frac{L}{8} \)
When the wire is plucked at a distance \( \frac{L}{8} \) from one end, the length of the vibrating segment is:
\[ L_1 = L - \frac{L}{8} = \frac{7L}{8} \]
For this length, the fundamental frequency (minimum frequency) can be expressed as:
\[ n = \frac{v}{\lambda} \]
The wavelength \( \lambda \) for the fundamental mode of vibration is twice the length of the vibrating segment:
\[ \lambda_1 = 2 \times \frac{7L}{8} = \frac{7L}{4} \]
Thus, substituting into the frequency formula, we get:
\[ n = \frac{v}{\frac{7L}{4}} = \frac{4v}{7L} \]

### Step 4: Second Plucking at \( \frac{L}{6} \)
Now, when the wire is plucked at a distance \( \frac{L}{6} \) from the other end, the length of the vibrating segment becomes:
\[ L_2 = L - \frac{L}{6} = \frac{5L}{6} \]
For this length, the wavelength for the fundamental mode of vibration is:
\[ \lambda_2 = 2 \times \frac{5L}{6} = \frac{5L}{3} \]
Now, we can find the new frequency \( n_2 \):
\[ n_2 = \frac{v}{\lambda_2} = \frac{v}{\frac{5L}{3}} = \frac{3v}{5L} \]

### Step 5: Relating the Frequencies
We know from the first case that:
\[ n = \frac{4v}{7L} \]
Now, we can express \( n_2 \) in terms of \( n \):
\[ n_2 = \frac{3v}{5L} = \left( \frac{3v}{5L} \right) \cdot \left( \frac{7L}{4v} \right) n = \frac{3 \cdot 7}{5 \cdot 4} n = \frac{21}{20} n \]

### Final Answer
Thus, the minimum frequency with which the wire vibrates when plucked at a distance \( \frac{L}{6} \) from the other end is:
\[ n_2 = \frac{21}{20} n \]