Two open organ pipes of length 50 , cm and 50 | vedclass

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(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L}$,where $v$ is the velocity of sound and $L$ is the length of the pip

Vedclass · Accepted Answer

$303$

Vedclass · Answer

(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L}$,where $v$ is the velocity of sound and $L$ is the length of the pipe.Given lengths are $L_1 = 50 \, cm = 0.5 \, m$ and $L_2 = 50.5 \, cm = 0.505 \, m$.The beat frequency is the difference between the fundamental frequencies: $f_1 - f_2 = 3 \, Hz$.Substituting the formula: $\frac{v}{2 	imes 0.5} - \frac{v}{2 	imes 0.505} = 3$.$v - \frac{v}{1.01} = 3$.$v \left( 1 - \frac{1}{1.01} ight) = 3$.$v \left( \frac{1.01 - 1}{1.01} ight) = 3$.$v \left( \frac{0.01}{1.01} ight) = 3$.$v = \frac{3 	imes 1.01}{0.01} = 3 	imes 101 = 303 \, m/s$.

Two open organ pipes of length $50 \, cm$ and $50.5 \, cm$ produce $3$ beats per second. The velocity of sound is ..... $m/s$.

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