Question

Two tuning forks \(A\) and \(B\) produce \(8\) beats in \(2\,\text{s}\) when sounded together. The frequency of tuning fork \(B\) is \(380\,\text{Hz}\). If tuning fork \(A\) is loaded with some wax, then they produce \(4\) beats in \(2\,\text{s}\). Find the original frequency of tuning fork \(A\).

• A. \(384\)
• B. \(388\)
• C. \(380\)
• D. \(392\)

Hint:

Adding wax to a tuning fork {always decreases its frequency}. Use this fact to eliminate incorrect solutions in beat problems.

Answer 1

The Correct Option is A

Concept:
Beat frequency is equal to the absolute difference of frequencies of two sources: \[ f_{\text{beats}} = |f_1 - f_2| \] Loading a tuning fork with wax reduces its frequency
.
Step 1: Determine Beat Frequencies
Initial beats: \[ \text{Beats per second} = \frac{8}{2} = 4\,\text{Hz} \] After loading wax: \[ \text{Beats per second} = \frac{4}{2} = 2\,\text{Hz} \]
Step 2: Find Possible Original Frequency of \(A\)
Let original frequency of tuning fork \(A = f_A\). From initial condition: \[ |f_A - 380| = 4 \] So, \[ f_A = 384\,\text{Hz} \quad \text{or} \quad 376\,\text{Hz} \]
Step 3: Use Effect of Wax Loading
After loading wax, frequency of \(A\) decreases
. Thus, the frequency difference with \(B\) becomes smaller: \[ |f_A' - 380| = 2 \] This is possible only if original frequency of \(A\) was greater than
\(380\,\text{Hz}\). Hence, \[ f_A = 384\,\text{Hz} \] \[ \boxed{f_A = 384\,\text{Hz}} \]