MCQ
Two sitar strings $A$ and $B$ playing the note $'Ga'$ are slightly out of tune and produce beats of frequency $6\,Hz$ . The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to $3\,Hz$. If the original frequency of $A$ is $324\,Hz$, what is the frequency of $B$ ..... $Hz$
  • A
    $324$
  • B
    $360$
  • C
    $330$
  • $318$

Answer

Correct option: D.
$318$
d
Frequency of string $\mathrm{A}, f_{A}=324 \mathrm{H} z$

Frequency of string $B=f_{B}$

Beat's frequency, $n=6 H z$

Beat's Frequency is given as$:$

$n=\left|f_{A}-f_{B}\right|$

$6=\left|324-f_{B}\right|$

$f_{B}=330 H z$ or $318 H z$

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as$:$

$f \propto \sqrt{T}$

Hence, the beat frequency cannot be $330 \mathrm{Hz}$

$\therefore f_{B}=318 H z$

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