Three similar wires of frequency `n_(1),n_(2)` and `n_(3)` are joined to make one wire. Its frequency will be

Options

A. `n=n_(1)+n_(2)+n_(3)`
B. `(1)/(n)=(1)/(n_1)+(1)/(n_2)+(1)/(n_3)`
C. `(1)/(sqrtn)=(1)/(sqrtn_1)+(1)/(sqrtn_2)+(1)/(sqrtn_3)`
D. `(1)/(n_2)=(1)/(n_1^2)+(1)/(n_2^2)+(1)/(n_3^2)`

Explanation

To find the frequency of the combined wire made from three similar wires with frequencies \( n_1, n_2, \) and \( n_3 \), we can follow these steps:

### Step 1: Understand the relationship between frequency, tension, and linear density
The frequency of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear density of the wire.

### Step 2: Establish the relationship for three wires
When three similar wires are joined, they can be treated as a single wire with a total length \( L = L_1 + L_2 + L_3 \). Since the wires are similar, they will have the same linear density \( \mu \) and tension \( T \).

### Step 3: Set up the equation for frequencies
For each wire, we can express the frequency as:
\[ n_1 = \frac{1}{2L_1} \sqrt{\frac{T}{\mu}}, \quad n_2 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}, \quad n_3 = \frac{1}{2L_3} \sqrt{\frac{T}{\mu}} \]

### Step 4: Relate the lengths and frequencies
From the above equations, we can express the lengths in terms of the frequencies:
\[ L_1 = \frac{1}{2n_1} \sqrt{\frac{T}{\mu}}, \quad L_2 = \frac{1}{2n_2} \sqrt{\frac{T}{\mu}}, \quad L_3 = \frac{1}{2n_3} \sqrt{\frac{T}{\mu}} \]

### Step 5: Substitute the lengths into the total length
The total length of the combined wire is:
\[ L = L_1 + L_2 + L_3 = \frac{1}{2n_1} \sqrt{\frac{T}{\mu}} + \frac{1}{2n_2} \sqrt{\frac{T}{\mu}} + \frac{1}{2n_3} \sqrt{\frac{T}{\mu}} \]
Factoring out \( \sqrt{\frac{T}{\mu}} \):
\[ L = \sqrt{\frac{T}{\mu}} \left( \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3} \right) \]

### Step 6: Find the frequency of the combined wire
The frequency of the combined wire can be expressed as:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Substituting the expression for \( L \):
\[ n = \frac{1}{2 \left( \sqrt{\frac{T}{\mu}} \left( \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3} \right) \right)} \sqrt{\frac{T}{\mu}} \]
This simplifies to:
\[ n = \frac{1}{\left( \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3} \right)} \]
Thus, the frequency of the combined wire is:
\[ n = \frac{1}{\frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}} \]

### Final Answer
The frequency of the combined wire is:
\[ n = \frac{1}{\frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}} \]