JEE Physics 2026 Question with Solution

Given:

• Original frequency of fork $B$ is $f_B = 380 \, \text{Hz}$
• Initial beats = $8$ in $2 \, \text{s}$
• After loading fork $A$ with wax, beats = $4$ in $2 \, \text{s}$

Find: The original frequency of tuning fork $A$.

Concept: Beat frequency is

$$f_b = |f_A - f_B|$$

Step 1: Initial beat frequency

$$\text{Beats per second} = \frac{8}{2} = 4$$

So,

$$|f_A - 380| = 4$$

Hence,

$$f_A = 384 \, \text{Hz} \quad \text{or} \quad 376 \, \text{Hz}$$

Step 2: Effect of loading with wax Loading a tuning fork decreases its frequency due to increase in effective mass.

After loading fork $A$ with wax, the beat frequency is reduced. That means the frequency difference between the two forks decreases. This is possible only if initially $f_A > f_B$, so that decreasing $f_A$ brings it closer to $f_B$.

Therefore,

$$f_A = 384 \, \text{Hz}$$

The original frequency of tuning fork $A$ is $384 \, \text{Hz}$.

From the initial beat frequency,

$$|f_A - 380| = 4$$

there are two possibilities:

$$f_A = 384 \, \text{Hz} \quad \text{or} \quad 376 \, \text{Hz}$$

If $f_A = 376 \, \text{Hz}$, then after loading with wax the frequency of fork $A$ would decrease further below $376 \, \text{Hz}$. Its difference from $380 \, \text{Hz}$ would then increase, so the beat frequency should increase, not reduce.

But the question states that the beat frequency reduces after loading. Therefore $376 \, \text{Hz}$ is not possible.

So the only valid value is $384 \, \text{Hz}$.