Two Sitar Strings A and B are Slightly Out Of Tune And Produce Beats Of Frequency 6 Hz. When The Tension In String A is Slightly Decreased, The Beat Frequency Is Found To Be Reduced To 3 Hz. If The Original Frequency Of A is 324 Hz, What Is The Frequency Of B?

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Question

Two sitar strings A and B are slightly out of tune and produce beats of frequency 6 Hz. When the tension in string A is slightly decreased, the beat frequency is found to be reduced to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Solution

Correct option is

318 Hz

 

  

The frequency of string B is   

              

Now, the frequency of a string is proportional to the square root of tension. Hence, if the tension in A is slightly decreased, its frequency will be slightly reduced, i.e. it will become less that 324 Hz. If the frequency of string B is 330 Hz, the beat frequency would increase to a value greater than 6 Hz if the tension in A is reduced. But the beat frequency is found to decrease to 3 Hz. Hence, frequency of B cannot be 330 Hz; it is, therefore 318 Hz. When the tension in A is reduced, its frequency becomes 324 – 3 = 321 Hz which will produce beats of frequency 3 Hz with string B of frequency 318 Hz.

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