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(B) Let $n_S$ be the frequency of the sonometer string and $n_f = 480 \ Hz$ be the frequency of the tuning fork.The beat frequency is given by $|n_f -

Vedclass · Accepted Answer

$470$

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(B) Let $n_S$ be the frequency of the sonometer string and $n_f = 480 \ Hz$ be the frequency of the tuning fork.The beat frequency is given by $|n_f - n_S| = 10 \ Hz$.This implies $n_S = 480 \pm 10$,so $n_S$ is either $470 \ Hz$ or $490 \ Hz$.The frequency of a sonometer string is given by $n_S = \frac{1}{2l} \sqrt{\frac{T}{m}}$. Thus,$n_S \propto \sqrt{T}$.When the tension $T$ is increased,$n_S$ increases.Case $1$: If $n_S = 470 \ Hz$,then $n_f - n_S = 10 \ Hz$. As $n_S$ increases,$n_f - n_S$ decreases (e.g.,if $n_S$ becomes $475 \ Hz$,beats become $5 \ Hz$). This matches the given condition.Case $2$: If $n_S = 490 \ Hz$,then $n_S - n_f = 10 \ Hz$. As $n_S$ increases,$n_S - n_f$ increases (e.g.,if $n_S$ becomes $495 \ Hz$,beats become $15 \ Hz$). This contradicts the given condition.Therefore,the frequency of the string must be $470 \ Hz$.

$A$ tuning fork of frequency $480 \ Hz$ produces $10$ beats per second when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before?

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