A tuning fork produces 4 beats per second whe | vedclass

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(A) The frequency of a sonometer wire is given by $f = \frac{v}{2L}$,where $v$ is constant as tension is constant. Thus,$f \propto \frac{1}{L}$,which

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$196$

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(A) The frequency of a sonometer wire is given by $f = \frac{v}{2L}$,where $v$ is constant as tension is constant. Thus,$f \propto \frac{1}{L}$,which means $fL = 	ext{constant}$.Let $f$ be the frequency of the tuning fork.For length $L_1 = 48 \ cm$,the wire frequency $f_1 = \frac{k}{48}$. Since it produces $4$ beats,$f_1 = f + 4$ or $f_1 = f - 4$. Since $L_1  f_2$,so $f_1 = f + 4$.For length $L_2 = 50 \ cm$,the wire frequency $f_2 = \frac{k}{50}$. Since it produces $4$ beats,$f_2 = f - 4$.Thus,$f_1 L_1 = f_2 L_2 \implies (f + 4) 	imes 48 = (f - 4) 	imes 50$.$48f + 192 = 50f - 200$.$2f = 392$.$f = 196 \ Hz$.

$A$ tuning fork produces $4$ beats per second when sounded with a sonometer wire of vibrating length $48 \ cm$. It also produces $4$ beats per second when the vibrating length is $50 \ cm$,keeping the tension in the wire the same. What is the frequency of the tuning fork in $Hz$?

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