1. Tardigrade
  2. Question
  3. Physics
  4. 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

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Q. 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

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Solution:

The frequency of string $A$, $\upsilon_{A}=427\,Hz$
Let original frequency of string $B$ be $\upsilon_{B}$
$\therefore \upsilon_{B}=\left(\upsilon_{A} \pm5\right)Hz=\left(427\pm5\right)Hz$
$= 432\, Hz$ or $422\, Hz$
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Increase in the tension of a string $B$, increase its frequency $(\upsilon\,\propto\,\sqrt{T})$
(i) If $\upsilon_{B}=432\,Hz$, 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 the question
$\therefore $ The original frequency of string $B$ be $422\, Hz$