Question

Two sitar strings A and B are slightly out of tune and produce beats of frequency 5 Hz. When the tension in the string B is slightly increased, the beat frequency is found to reduce to 3 Hz. If the frequency of string A is 427 Hz, the original frequency of string B is
A. 422 Hz
B. 424 Hz
C. 430 Hz
D. 432 Hz

Answer 1

Correct Answer - A
The frequency of string A, `upsilon_(A)= 427 Hz`
Let original frequency of string B be `upsilon_(B)`
`thereforeupsilon_(B)=(upsilon_(A)pm5)Hz=(427pm5)Hz`
`=432Hzor422Hz`

Increase the tension of a string B, increase its frequency `(vpropsqrtT)`
(i) `upsilon_(B)`=432 Hz, a further increase in `upsilon_(B)`, increase the beat frequency. But this is not given in the question.
(ii) If `upsilon_(B)=422` Hz, a further increase in `upsilon_(B)`, decrease the beat frequency. This is given in this question.
`therefore` The original frequency of string B be 422 Hz