5, text{beats/second} are heard when a tuning | vedclass

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(A) Let the frequency of the tuning fork be $n$ and the frequency of the sonometer wire be $f$. Since the beat frequency is $5\, 	ext{Hz}$, the frequ

Vedclass · Accepted Answer

$195$

Vedclass · Answer

(A) Let the frequency of the tuning fork be $n$ and the frequency of the sonometer wire be $f$. Since the beat frequency is $5\, 	ext{Hz}$, the frequency of the wire is either $(n + 5)$ or $(n - 5)$. For a sonometer wire, frequency $f \propto \frac{1}{l}$, which implies $f_1 l_1 = f_2 l_2$. Since $l_1 = 0.95\, 	ext{m}$ and $l_2 = 1.0\, 	ext{m}$, the frequency at $0.95\, 	ext{m}$ is higher than at $1.0\, 	ext{m}$. Thus, we have $(n + 5) 	imes 0.95 = (n - 5) 	imes 1.0$. $0.95n + 4.75 = n - 5$. $0.05n = 9.75$. $n = \frac{9.75}{0.05} = 195\, 	ext{Hz}$.

$5\, \text{beats/second}$ are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95\, \text{m}$ or $1\, \text{m}$. The frequency of the fork will be ... $\text{Hz}$.

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