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

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(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$,where $v$ is the speed of sound and $L$ is the length of the

Vedclass · Accepted Answer

$308 \, Hz, 300 \, Hz$

Vedclass · Answer

(A) For a pipe closed at one end,the fundamental frequency is given by $n = \frac{v}{4L}$,where $v$ is the speed of sound and $L$ is the length of the air column.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 $\frac{v}{4 	imes 37.5 	imes 10^{-2}} - \frac{v}{4 	imes 38.5 	imes 10^{-2}} = 8$.$\frac{v}{4 	imes 10^{-2}} \left( \frac{1}{37.5} - \frac{1}{38.5} ight) = 8$.$\frac{v}{0.04} \left( \frac{38.5 - 37.5}{37.5 	imes 38.5} ight) = 8$.$\frac{v}{0.04} \left( \frac{1}{1443.75} ight) = 8 \implies v = 8 	imes 0.04 	imes 1443.75 = 462 \, m/s$.Now,$n_A = \frac{462}{4 	imes 0.375} = \frac{462}{1.5} = 308 \, Hz$.$n_B = n_A - 8 = 308 - 8 = 300 \, Hz$.

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