Answer
Verified
97.8k+ views
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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
Calculate CFSE of the following complex FeCN64 A 04Delta class 11 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A mosquito with 8 legs stands on the water surface class 11 physics JEE_Main
A ball of mass 05 Kg moving with a velocity of 2ms class 11 physics JEE_Main
The focal length of a thin biconvex lens is 20cm When class 12 physics JEE_Main