Equal forces are applied to stretch two wires of same material and same diameters. The ratio of their lengths is 3 : 4. If the first wire has a frequency of 512 Hz, then calculate the frequency of other wire.

Answer

To solve the problem, we need to find the frequency of the second wire given the frequency of the first wire and the ratio of their lengths. Let's break it down step by step.

### Step 1: Understand the relationship between frequency, length, and tension
The frequency of a vibrating wire can be expressed using 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 mass per unit length (linear density) of the wire.

### Step 2: Identify the given values
From the problem:
- The lengths of the wires are in the ratio \( L_1 : L_2 = 3 : 4 \).
- The frequency of the first wire \( f_1 = 512 \, \text{Hz} \).
- Both wires are made of the same material and have the same diameter, which means they have the same tension \( T \) and mass per unit length \( \mu \).

### Step 3: Write the frequency equations for both wires
Using the formula for frequency:
- For the first wire:
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T}{\mu}} \]
- For the second wire:
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}} \]

### Step 4: Set up the ratio of the frequencies
Taking the ratio of the two frequencies:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \]
This is because \( \sqrt{\frac{T}{\mu}} \) cancels out since it is the same for both wires.

### Step 5: Substitute the lengths in terms of the ratio
Given the lengths:
- Let \( L_1 = 3x \)
- Let \( L_2 = 4x \)

Now substituting these values into the frequency ratio:
\[ \frac{f_1}{f_2} = \frac{4x}{3x} = \frac{4}{3} \]

### Step 6: Solve for \( f_2 \)
Rearranging the equation gives:
\[ f_2 = f_1 \cdot \frac{3}{4} \]
Substituting the value of \( f_1 \):
\[ f_2 = 512 \cdot \frac{3}{4} \]

### Step 7: Calculate \( f_2 \)
Now, calculate \( f_2 \):
\[ f_2 = 512 \cdot 0.75 = 384 \, \text{Hz} \]

### Conclusion
The frequency of the second wire is \( 384 \, \text{Hz} \).

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