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Question: When a guitar string is sounded with a \(440Hz\) tuning fork, a beat frequency of \(5Hz\) is heard. ...

When a guitar string is sounded with a 440Hz440Hz tuning fork, a beat frequency of 5Hz5Hz is heard. If the experiment is repeated with a tuning fork of 437Hz437Hz, the beat frequency is 8Hz8Hz. The string frequency(Hz)\left( {Hz} \right) is
A. 445445
B. 435435
C. 429429
D. 448448

Explanation

Solution

When two waves of nearly same frequencies interfere with each other and produce maximum or minimum intensity is called as beats. Consider the beats produced by the combinations of the guitar with the two tuning forks, apply the definition of beats to find the frequency of the guitar string.

Complete step by step answer:
One beat is one cycle of maximum and minimum intensities, meaning that the frequency of beats is given by the difference in the individual frequencies of the two waves. We have the frequency of one tuning fork as 440Hz440Hz and frequency of the other tuning fork as 437Hz437Hz. Given that the beats heard when the first tuning fork is sounded with the guitar string is 5Hz5Hz, mathematically we have,
fs440=5 fs440=±5  \left| {{f_s} - 440} \right| = 5 \\\ \Rightarrow{f_s} - 440 = \pm 5 \\\
We can say that the frequency of the guitar string will be either fs=445{f_s} = 445 or fs=435{f_s} = 435.

Now, if we sound the guitar string with the tuning fork of frequency 437Hz437Hzthen the beats that are supposed to be heard have to be equal to 88beats. If the frequency of the guitar string was 445Hz445Hz, the beats produced will be equal to 445437=8Hz445 - 437 = 8Hz which satisfies our condition.

Let us check for the other case. If the frequency of the guitar string was 435Hz435Hz, the beats produced will be 437435=2Hz437 - 435 = 2Hz. As you can see that this does not satisfy our condition and hence, frequency of 435Hz435Hz is rejected. Therefore, the string frequency(Hz)\left( {Hz} \right) is 445445.

Hence,option A is correct.

Note: The beat frequency is given by mode of the difference between the individual frequencies. Mathematically, beats heard per second are given as f1f2\left| {{f_1} - {f_2}} \right|. Since we knew the frequencies of the tuning fork, we did not consider the mode sign. This question could have been asked in a different way. Such as, if the tuning fork was loaded with wax and with some conditions given, find out the frequency. In this case, remember that whenever a tuning fork is loaded with wax, its frequency will decrease. By applying the same analysis, you could have got the same correct answer.