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Question

(A) For a sonometer wire,the frequency $f$ is inversely proportional to the length $l$ of the wire,i.e.,$f \propto 1/l$ or $f \cdot l = k$ (constant).

Vedclass · Accepted Answer

$96 \ Hz, 92 \ Hz$

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(A) For a sonometer wire,the frequency $f$ is inversely proportional to the length $l$ of the wire,i.e.,$f \propto 1/l$ or $f \cdot l = k$ (constant).Let the frequencies of the two tuning forks be $f_1$ and $f_2$.Given $f_1 \cdot 23 = f_2 \cdot 24 = k$.This implies $f_1 = k/23$ and $f_2 = k/24$.Since $f_1 > f_2$,the beat frequency is $f_1 - f_2 = 4$.Substituting the values: $k/23 - k/24 = 4$.Taking $k$ as common: $k(24 - 23) / (23 \cdot 24) = 4$.$k / 552 = 4 \implies k = 2208$.Now,$f_1 = 2208 / 23 = 96 \ Hz$.And $f_2 = 2208 / 24 = 92 \ Hz$.Thus,the frequencies are $96 \ Hz$ and $92 \ Hz$.

Two tuning forks when sounded together produce $4$ beats per second. One of the forks is in unison with $23 \ cm$ length of a sonometer wire and the other with $24 \ cm$ length of the same wire. The frequencies of the two tuning forks are

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