Two tuning forks A and B produce 8 , Hz beats | vedclass

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Question

(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$.For tuning fork $A$,$n_A = \frac{v}{4 	imes 37.5 	imes 10^

Vedclass · Accepted Answer

$308 \, Hz, 300 \, Hz$

Vedclass · Answer

(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$.For tuning fork $A$,$n_A = \frac{v}{4 	imes 37.5 	imes 10^{-2}}$.For tuning fork $B$,$n_B = \frac{v}{4 	imes 38.5 	imes 10^{-2}}$.Since $n_A > n_B$ (as $L_A $\frac{v}{4 	imes 0.375} - \frac{v}{4 	imes 0.385} = 8$.$\frac{v}{4} \left( \frac{38.5 - 37.5}{37.5 	imes 38.5} 	imes 100 ight) = 8$.$\frac{v}{4} \left( \frac{1}{1443.75} 	imes 100 ight) = 8 \implies v = \frac{8 	imes 4 	imes 1443.75}{100} = 462 \, m/s$.Now,$n_A = \frac{462}{4 	imes 0.375} = 308 \, Hz$.$n_B = n_A - 8 = 300 \, Hz$.

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