Question

The frequency of two tuning forks A and B are 1.5% more and 2.5% less than that of the tuning fork C. When A and B are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork C is

• A. 200 Hz
• B. 300 Hz
• C. 240 Hz
• D. 360 Hz

Hint:

When two tuning forks produce beats, the beat frequency is the absolute difference between their frequencies. Use the given beat frequency to calculate the unknown frequency.

Answer 1

The Correct Option is B

Step 1: Understanding the beat frequency.
The beat frequency \( f_{beats} \) is given by the difference in frequencies of the two tuning forks: \[ f_{beats} = |f_A - f_B| \] We are given that 12 beats are produced in 1 second, so the frequency difference between A and B is 12 Hz.

Step 2: Frequency relation between A, B, and C.
Let the frequency of tuning fork C be \( f_C \). Then, - The frequency of A is \( f_A = 1.015 f_C \), - The frequency of B is \( f_B = 0.975 f_C \).
The difference in frequencies between A and B is: \[ f_A - f_B = 1.015 f_C - 0.975 f_C = 0.04 f_C \] We are given that the beat frequency is 12 Hz, so: \[ 0.04 f_C = 12 \quad \Rightarrow \quad f_C = \frac{12}{0.04} = 300 \, \text{Hz} \]
Step 3: Conclusion.
The frequency of tuning fork C is 300 Hz. Thus, the correct answer is (B).