Question

two sitar strings $A$ and $B$ are slightly out of tune and produces beat of frequency $5Hz$. When the tension in the string $B$ is slightly increased, the beat frequency is found to reduce to $3Hz$. If the frequency of string $A$ is $472Hz$, the original frequency of string $B$ is

• A. $422Hz$
• B. $424Hz$
• C. $430Hz$
• D. $432Hz$

Answer 1

The correct option is A $422Hz$

The frequency of string $A,{\upsilon }_A=427Hz$

Let the original frequency of string $B$ be ${\upsilon }_B$.

$\therefore {\upsilon }_B=({\upsilon }_A\pm 5)Hz(427\pm 5)Hz$

$=432Hz$ or $422Hz$

Increase in the tension of a string $B$, increase its frequency $\left(\upsilon \alpha \sqrt{T}\right)$

(i) If ${\upsilon }_B=432Hz$, a further increase in ${\upsilon }_B$, increase the beat frequency. But this is not given in the question.

(ii) If ${\upsilon }_B=422Hz$, a further increase in ${\upsilon }_B$, decrease the beat frequency. This is given in the frequency question.

$\therefore$ The original frequency of string $B$ be $422Hz$.