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

Answer: (A)

422 Hz

Answer 2

The frequency of string A,

$\upsilon_{A} = 427Hz$

Let original frequency of string B be $\upsilon_{B}$.

$\therefore\upsilon_{B} = (\upsilon_{A} \pm 5)Hz = (427 \pm 5)Hz$

= 432 Hz or 422 Hz

Increase in the tension of a string, B increase its frequency

$(\upsilon \propto \sqrt{T})$

(i) If $\upsilon_{B} = 432Hz,$ a further increase in $\upsilon_{B},$

increases the beat frequency. But this is not given in the questions.

(ii) If $\upsilon_{B} = 422Hz,$ a further increase in $\mathrm { U } _ { \mathrm { B } }$

decreases the beat frequency. This is given in the equations. $\therefore$The original frequency of strings B is 422 Hz.