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# When two tuning forks A and B are sounded together, x beats per seconds are heard. Frequency of A is n. Now when one pong of B is loaded with wax, the number of beats per seconds decreases. The frequency of B is:A. n-xB. n+xC. n+2xD. n+x.x

Last updated date: 20th Jun 2024
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Hint: Beats are the interference pattern between two sources of slightly different frequencies. Beats can only be observed by human ears when the two sources have almost the same frequency. Coming to frequency, it’s the property of the source producing the waves. Now if the source is waxed, that means we are applying wax or putting some weight onto the forks. Hence the frequency of the fork will decrease due to increase in inertia.

Formula used: $Beat\ frequency (\beta)=|\nu_1-\nu_2|$

Since initially the beat frequency was x, and the frequency of A is ‘n’, hence putting in the equation, we get
$\beta=|\nu_1-\nu_2|$
Or $x=|n-\nu_2|$
But, to open the mathematical sign mode, we need to know which fork has more frequency? We can easily figure it out by analyzing the statement in the question.
According to the statement, when fork B is loaded with wax, the beat frequency decreases.
This can only be possible if the frequency of fork B is greater than that of A.
Thus, $x=\nu_{B}-n$
Or $\nu_{B}=n+x$
[To understand properly, see note]

So, the correct answer is “Option B”.

Note: As loading B will decrease the beat frequency, that means $\beta$ decrease on decreasing the frequency of fork B. This means that beat frequency and frequency of B must be directly related.
$Beat\ frequency (\beta)=\nu_1-\nu_2$
Let’s assume $\nu_2$be the frequency of B, then on decreasing $\nu_2$, $\beta$ will definitely increase. Hence our assumption is wrong. Whereas assuming $\nu_1$ being the frequency of B, then on decreasing $\nu_1$, $\beta$ will also decrease.