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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 Hz
B) 514 Hz
C) 516 Hz
D) 508 Hz

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Last updated date: 20th Jun 2024
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Answer
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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.

Complete answer:
We know that the number of beats per second, or the beat frequency, is the difference between two frequencies.
We have a tuning fork of frequency $\nu_{fork} = 512\;Hz$.
We are told that it makes 4 beats per second with the string of the piano. This means that:
$\nu_{piano} = \nu_{fork} \pm 4$.
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.
Therefore, the frequency of the piano string before increasing the tension is given by the option D 508 Hz.

Note: This calculation is not just confined to the just calculations as this also has an application in real life. We can use the above method to determine an unknown frequency. Suppose you have a tuning fork A of unknown frequency, and you have a tonometer consisting of tuning forks of known frequencies. Sound the fork A with each of the tuning forks in the tonometer until you find a tonometer fork that produces countable beats per second with the unknown fork. Since the number of beats is the difference between the tonometer fork and the unknown fork frequencies, you can find the frequency of the unknown fork this way. This is also the same way musical instruments are tuned, except you strive to not hear any beats in this case and they are sounded in unison.