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# A tuning fork of frequency $512\;Hz$ makes 4 beats per second with the vibrating string of a string of a piano. The beat frequency decreases to 2 beats per second, when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was:A) 510 HzB) 514 HzC) 516 HzD) 508 Hz

Last updated date: 11th Sep 2024
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Hint: Recall that the beat frequency is nothing but the difference in the frequencies of the two progressive waves. Using this, we get two possible piano frequencies. See which one you can eliminate given that, if the frequency of the piano is increased, then the beat frequency decreases. The frequency that suits this criteria will be the piano frequency before tightening its string.

Formula used: Beat frequency $= \nu_1 - \nu_2$, where $\nu_1$ and $\nu_2$ are the two frequencies whose propagation causes a beat.

We have a tuning fork of frequency $\nu_{fork} = 512\;Hz$.
$\nu_{piano} = \nu_{fork} \pm 4$.
Now, when the tension in the piano string is increased, this means that the $\nu_{piano}$ will also increase, and we are given that the beat frequency decreases to 2 beats per second.
If $\nu_{piano} = 512 +4 =516\;Hz$ , increasing the piano frequency further would lead to an increase in beat frequency. But this is not the case.
So, if $\nu_{piano} = 512 -4 =508\;Hz$, increasing the piano frequency further will lead to a decrease in beat frequency. This is what happens, as given in the question.