Question

Two tuning forks A and B produce 10 beats per second. On loading a small ring on one prong of B again 10 beats per second are produced. What was the frequency of B before loading small ring if now frequency of B is 430 Hz. Give reason for your answer.​

Answer 1

Explanation:

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Answer 2

Given:

The number of beats per second produced by forks A and B = 10

The number of beats per second produced by forks A and B  after loading B = 10

The  frequency of B after loading = 430 Hz

To Find:

The frequency of B before loading small ring

Solution:

The number of beats per second produced by 2 forks = $| f_a - f_b|$ where $f_a$ and $f_b$ are the frequencies of forks A and B.

According to the question,

$|f_a - f_b| = 10$          - (1)

When a ring is loaded on fork B, its frequency decreases. Let the new frequency be $f'_b$ = 430 Hz

Since even on decreasing the frequency of fork, the difference in frequency of the two forks remain the same

⇒ $f_b > f_a > f'_b$

$f_a - f'_b = 10$

or $f_a$ = 10 + 430

= 440 Hz

Substituting in equation 1,

$f_b$ = 440 + 10

= 450 Hz

Hence, the frequency of B before loading the small ring was 450 Hz.