Fundamental frequency of sonometer wire is n. If the length, tension and diameter of wire are tripled the new fundamental frequency is

Options

A. `(n)/sqrt3`
B. `n/3`
C. `nsqrt3`
D. `(n)/(3sqrt3)`

Answer

To find the new fundamental frequency of the sonometer wire when the length, tension, and diameter are all tripled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Fundamental Frequency**:
The fundamental frequency \( f \) of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

2. **Express Mass per Unit Length**:
The mass per unit length \( \mu \) can be expressed in terms of the diameter \( d \) (or radius \( r \)) and density \( \rho \):
\[ \mu = \pi r^2 \rho \]
Since diameter \( d = 2r \), we can also express it as:
\[ \mu = \frac{\pi d^2}{4} \rho \]

3. **Identify Changes in Parameters**:
According to the problem, the length \( L \), tension \( T \), and diameter \( d \) of the wire are all tripled:
- New length \( L_2 = 3L \)
- New tension \( T_2 = 3T \)
- New diameter \( d_2 = 3d \)

4. **Calculate New Mass per Unit Length**:
Since the diameter is tripled, the new mass per unit length \( \mu_2 \) becomes:
\[ \mu_2 = \pi \left(\frac{d_2}{2}\right)^2 \rho = \pi \left(\frac{3d}{2}\right)^2 \rho = \pi \frac{9d^2}{4} \rho = 9 \mu \]

5. **Substitute New Values into the Frequency Formula**:
Now, substituting the new values into the frequency formula:
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu_2}} = \frac{1}{2(3L)} \sqrt{\frac{3T}{9\mu}} = \frac{1}{6L} \sqrt{\frac{3T}{\mu}} \]

6. **Relate New Frequency to Original Frequency**:
The original frequency \( f_1 \) is:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Therefore, we can express the new frequency \( f_2 \) in terms of \( f_1 \):
\[ f_2 = \frac{1}{6} \cdot 3 \cdot f_1 = \frac{1}{2\sqrt{3}} f_1 \]
Since \( f_1 = n \), we have:
\[ f_2 = \frac{n}{3\sqrt{3}} \]

7. **Final Result**:
Thus, the new fundamental frequency is:
\[ f_2 = \frac{n}{3\sqrt{3}} \]

### Conclusion:
The new fundamental frequency of the sonometer wire when the length, tension, and diameter are tripled is:
\[ \boxed{\frac{n}{3\sqrt{3}}} \]