When two tuning forks are sounded together,6 | vedclass

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(B) For a sonometer wire,the frequency $f$ is inversely proportional to the vibrating length $l$,i.e.,$f \propto 1/l$. Let the frequencies of the two

Vedclass · Accepted Answer

$414 \ Hz, 420 \ Hz$

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(B) For a sonometer wire,the frequency $f$ is inversely proportional to the vibrating length $l$,i.e.,$f \propto 1/l$. Let the frequencies of the two tuning forks be $f_1$ and $f_2$. Given that $f_1 \propto 1/0.70$ and $f_2 \propto 1/0.69$. Since $f_2 > f_1$,we have $f_2 - f_1 = 6 \ Hz$. Let $f_1 = k/0.70$ and $f_2 = k/0.69$,where $k$ is a constant. Substituting these into the beat frequency equation: $k/0.69 - k/0.70 = 6$. $k(0.70 - 0.69) / (0.69 	imes 0.70) = 6$. $k(0.01) / 0.483 = 6$. $k = 6 	imes 0.483 / 0.01 = 6 	imes 48.3 = 289.8$. Now,$f_1 = 289.8 / 0.70 = 414 \ Hz$. And $f_2 = 289.8 / 0.69 = 420 \ Hz$. Thus,the frequencies are $414 \ Hz$ and $420 \ Hz$.

When two tuning forks are sounded together,$6$ beats per second are heard. One of the forks is in unison with $0.70 \ m$ length of a sonometer wire and another fork is in unison with $0.69 \ m$ length of the same sonometer wire. The frequencies of the two tuning forks are:

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