Two tuning forks A and B produce 8 Hz beats p | 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 $n_A - n_B = 8\ Hz$,we have $\frac{v}{4 	imes 37.5 	imes 10^{-2}} - \frac{v}{4 	imes 38.5 	imes 10^{-2}} = 8$.$\frac{v}{4 	imes 10^{-2}} \left( \frac{1}{37.5} - \frac{1}{38.5} ight) = 8$.$\frac{v}{0.04} \left( \frac{38.5 - 37.5}{37.5 	imes 38.5} ight) = 8$.$\frac{v}{0.04} \left( \frac{1}{1443.75} ight) = 8 \Rightarrow v = 8 	imes 0.04 	imes 1443.75 = 462\ m/s$.Now,$n_A = \frac{462}{4 	imes 0.375} = 308\ Hz$.And $n_B = n_A - 8 = 300\ Hz$.

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