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:

Let nAandnBnAandnB be the frequencies of fork A and B respectively. There are two possibilities

nA>nBornA<nBnA>nBornA<nB

If nA>nBnA>nB

i.e. nA−nB=10nA-nB=10

nA=nB+10=480+10=490HznA=nB+10=480+10=490Hz

On loading fork A its frequency decreases in other words nAnA decreases.

∴nA−nB=10∴nA-nB=10 also decreases.

But here the beat frequency increases to 15 beats.

Hence nAnA is not greater than n2n2.

i.e. nA<nBnA<nB in this case nB−nA=10nB-nA=10

nA=nB−10=480−10=470Hz

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hope it helps