Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is

Q. Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is ______ Hz.

Correct Answer: 384

Solution

Beat frequency is equal to the absolute difference of frequencies of the two tuning forks.

Initial beat frequency = 8 / 2 = 4 Hz

Let the original frequency of tuning fork A be f. Given frequency of tuning fork B is 380 Hz.

|f − 380| = 4

So possible values of f are 384 Hz or 376 Hz.

When wax is loaded on fork A, its frequency decreases. New beat frequency:

4 / 2 = 2 Hz

This means the difference between frequencies has reduced, so original frequency of A must be greater than 380 Hz.

Therefore, f = 384 Hz

Hence, the original frequency of tuning fork A is 384 Hz.

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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