A tuning fork vibrating with a sonometer havi | vedclass

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(C) Let the frequency of the tuning fork be $f$ and the frequency of the sonometer wire be $n = \frac{v}{2L}$.Given that the beat frequency is $5 \,Hz

Vedclass · Accepted Answer

$205$

Vedclass · Answer

(C) Let the frequency of the tuning fork be $f$ and the frequency of the sonometer wire be $n = \frac{v}{2L}$.Given that the beat frequency is $5 \,Hz$,we have $|f - n| = 5$.For $L_1 = 0.20 \,m$,$n_1 = \frac{v}{2 	imes 0.20} = \frac{v}{0.4}$.For $L_2 = 0.21 \,m$,$n_2 = \frac{v}{2 	imes 0.21} = \frac{v}{0.42}$.Since $L_2 > L_1$,$n_2 If the beat frequency remains $5 \,Hz$,the tuning fork frequency $f$ must lie between $n_1$ and $n_2$ or be such that the difference remains constant. Case $1$: $n_1 - f = 5 \implies f = n_1 - 5$.Case $2$: $f - n_2 = 5 \implies f = n_2 + 5$.Equating the two: $n_1 - 5 = n_2 + 5 \implies n_1 - n_2 = 10$.Substituting $n_1 = \frac{v}{0.4}$ and $n_2 = \frac{v}{0.42}$:$\frac{v}{0.4} - \frac{v}{0.42} = 10 \implies v \left( \frac{0.42 - 0.4}{0.168} ight) = 10 \implies v \left( \frac{0.02}{0.168} ight) = 10 \implies v = \frac{1.68}{0.02} = 84 \,m/s$.Now,$n_1 = \frac{84}{0.4} = 210 \,Hz$.Since $f = n_1 - 5 = 210 - 5 = 205 \,Hz$.

$A$ tuning fork vibrating with a sonometer having a wire of length $20 \,cm$ produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \,cm$. The frequency of the tuning fork must be ............ $Hz$.

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