Question

Two sitar strings $A$ and $B$ are slightly out of tune and produce beats 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 $427Hz$, the original frequency of string $B$ is

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

Answer 1

Answer: (A)

The frequency of string $A$, ${\upsilon }_A=427Hz$
Let original frequency of string $B$ be ${\upsilon }_B$
$\therefore {\upsilon }_B=\left({\upsilon }_A\pm 5\right)Hz=\left(427\pm 5\right)Hz$
$=432Hz$ or $422Hz$

Increase in the tension of a string $B$, increase its frequency $(\upsilon \propto \sqrt{T})$
(i) If ${\upsilon }_B=432Hz$, a further increase in ${\upsilon }_B$, 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 question
$\therefore$ The original frequency of string $B$ be $422Hz$