Question

Two identical wires are vibrating in unison. If the tension in one of the wires is increased by 2%, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is

• A. 1000 Hz
• B. 500 Hz
• C. 400 Hz
• D. 200 Hz

Hint:

When the tension in a vibrating wire is changed, the frequency changes according to the square root of the tension. Use this to calculate the change in frequency and the resulting beat frequency.

Answer 1

The Correct Option is B

Step 1: Relationship between frequency and tension.
The frequency of a vibrating wire is proportional to the square root of the tension, i.e., \[ f \propto \sqrt{T} \] Let the initial frequency be \( f_0 \) and the initial tension be \( T_0 \). If the tension is increased by 2%, the new frequency will be \( f_1 \), and we have: \[ \frac{f_1}{f_0} = \sqrt{\frac{T_1}{T_0}} = \sqrt{1.02} \] This gives a new frequency \( f_1 = f_0 \times \sqrt{1.02} \). The beat frequency is given by the difference between the two frequencies: \[ \text{Beat frequency} = |f_1 - f_0| \] We are given that the beat frequency is 5 Hz. Therefore, \[ |f_1 - f_0| = 5 \quad \Rightarrow \quad |f_0 \times (\sqrt{1.02} - 1)| = 5 \]
Step 2: Solving for the initial frequency.
Using the approximation \( \sqrt{1.02} \approx 1 + 0.01 \), we get: \[ |f_0 \times (1.01 - 1)| = 5 \quad \Rightarrow \quad f_0 \times 0.01 = 5 \quad \Rightarrow \quad f_0 = 500 \, \text{Hz} \]
Step 3: Conclusion.
The initial frequency of each wire is 500 Hz, which corresponds to option (B).