Question

A tuning fork \(P\) of frequency \(384\,\mathrm{Hz}\) produces 6 beats per second with a tuning fork \(Q\). When a little wax is attached to \(Q\) and again \(P\) and \(Q\) are sounded together, the number of beats produced do not change. The initial frequency of \(Q\) is

• A. \(390\,\mathrm{Hz}\)
• B. \(378\,\mathrm{Hz}\)
• C. \(381\,\mathrm{Hz}\)
• D. \(387\,\mathrm{Hz}\)

Hint:

Beat frequency is \[ \boxed{ f_b=|f_1-f_2| } \] Attaching wax to a tuning fork decreases its natural frequency.

Answer 1

The Correct Option is A

Step 1: Determine the possible frequencies of \(Q\). Since 6 beats are heard, \[ |f_Q-384|=6. \] Hence, \[ f_Q=390~\text{Hz} \quad\text{or}\quad 378~\text{Hz}. \]

Step 2: Use the effect of attaching wax. Attaching wax decreases the frequency of a tuning fork. If \[ f_Q=378~\text{Hz}, \] its frequency becomes still smaller, so the beat frequency would become greater than \(6\), which contradicts the question. If \[ f_Q=390~\text{Hz}, \] its frequency decreases toward \(384\,\text{Hz}\). Since the beats remain \(6\), the frequency must still be above \(384\,\text{Hz}\), which is possible. Therefore, \[ \boxed{f_Q=390~\text{Hz}.} \] Hence, \[ \boxed{(A)} \] is the correct answer.