Two wires of the same material and same cross section are stretched on a sonometer. One wire is loaded with 1 kg and another is loaded with 9 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire.

Answer

To solve the problem, we need to use the relationship between the fundamental frequency of vibration, tension, and vibrating length of the wires. Let's denote the variables as follows:

- \( L_1 \): vibrating length of the first wire = 60 cm = 0.6 m
- \( T_1 \): tension in the first wire = mass × g = 1 kg × 9.8 m/s² = 9.8 N
- \( L_2 \): vibrating length of the second wire (unknown)
- \( T_2 \): tension in the second wire = mass × g = 9 kg × 9.8 m/s² = 88.2 N

Since the fundamental frequencies of both wires are the same, we can set up the equation based on the formula for fundamental frequency:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Where:
- \( f \) is the fundamental frequency
- \( L \) is the vibrating length
- \( T \) is the tension
- \( \mu \) is the mass per unit length

Since both wires are made of the same material and have the same cross-sectional area, their mass per unit length \( \mu \) is the same. Therefore, we can write:

\[ \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}} = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}} \]

The \( \mu \) and \( \frac{1}{2} \) cancel out, leading to:

\[ \frac{1}{L_1} \sqrt{T_1} = \frac{1}{L_2} \sqrt{T_2} \]

Rearranging gives us:

\[ L_2 = L_1 \cdot \frac{\sqrt{T_2}}{\sqrt{T_1}} \]

Now, substituting the known values:

1. Calculate \( T_1 \) and \( T_2 \):
- \( T_1 = 9.8 \, \text{N} \)
- \( T_2 = 9 \times 9.8 = 88.2 \, \text{N} \)

2. Substitute \( L_1 \), \( T_1 \), and \( T_2 \) into the equation for \( L_2 \):

\[ L_2 = 0.6 \cdot \frac{\sqrt{88.2}}{\sqrt{9.8}} \]

3. Calculate \( \sqrt{88.2} \) and \( \sqrt{9.8} \):
- \( \sqrt{88.2} \approx 9.38 \)
- \( \sqrt{9.8} \approx 3.13 \)

4. Now substitute these values back into the equation for \( L_2 \):

\[ L_2 = 0.6 \cdot \frac{9.38}{3.13} \approx 0.6 \cdot 3 = 1.8 \, \text{m} \]

Thus, the vibrating length of the second wire \( L_2 \) is approximately **1.8 m**.