Two different stretched wires have same tension and mass per unit length. Fifth overtone frequency of the first wire is equal to second harmonic frequency of the second wire. Find the ratio of their lengths.

Explanation

To solve the problem, we need to find the ratio of the lengths of two different stretched wires based on the given conditions. Let's break down the solution step by step.

### Step 1: Understand the Frequencies
We know that:
- The frequency of a wire is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the velocity of the wave in the wire, and \( L \) is the length of the wire.

### Step 2: Velocity of the Waves
The velocity \( V \) of a wave in a stretched wire can be expressed as:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, and \( \mu \) is the mass per unit length of the wire.

### Step 3: Fifth Overtone Frequency of the First Wire
The fifth overtone corresponds to the sixth harmonic. Therefore, the frequency of the first wire can be expressed as:
\[ f_1 = 6f_0 = 6 \cdot \frac{V_1}{2L_1} = \frac{6V_1}{2L_1} = \frac{3V_1}{L_1} \]
Substituting for \( V_1 \):
\[ f_1 = \frac{3}{L_1} \sqrt{\frac{T}{\mu}} \]

### Step 4: Second Harmonic Frequency of the Second Wire
The second harmonic frequency of the second wire is given by:
\[ f_2 = 2f_0 = 2 \cdot \frac{V_2}{2L_2} = \frac{V_2}{L_2} \]
Substituting for \( V_2 \):
\[ f_2 = \frac{1}{L_2} \sqrt{\frac{T}{\mu}} \]

### Step 5: Equate the Frequencies
According to the problem, the fifth overtone frequency of the first wire is equal to the second harmonic frequency of the second wire:
\[ \frac{3}{L_1} \sqrt{\frac{T}{\mu}} = \frac{1}{L_2} \sqrt{\frac{T}{\mu}} \]

### Step 6: Simplify the Equation
Since both sides contain \( \sqrt{\frac{T}{\mu}} \), we can cancel it out:
\[ \frac{3}{L_1} = \frac{1}{L_2} \]

### Step 7: Find the Ratio of Lengths
Rearranging the equation gives:
\[ \frac{L_1}{L_2} = 3 \]
Thus, the ratio of the lengths of the two wires is:
\[ L_1 : L_2 = 3 : 1 \]

### Final Answer
The ratio of the lengths of the two wires is \( L_1 : L_2 = 3 : 1 \).

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