Two open organ pipes of lengths 50 ,cm and 51 | vedclass

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(D) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2l}$.Given, beat frequency $f_b = \frac{40}{10} = 4 \,Hz$.Let $l_1 = 50

Vedclass · Accepted Answer

$204$

Vedclass · Answer

(D) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2l}$.Given, beat frequency $f_b = \frac{40}{10} = 4 \,Hz$.Let $l_1 = 50 \,cm = 0.5 \,m$ and $l_2 = 51 \,cm = 0.51 \,m$.The difference in frequencies is $f_1 - f_2 = 4 \,Hz$.$\frac{v}{2l_1} - \frac{v}{2l_2} = 4$$\frac{v}{2} \left( \frac{1}{0.5} - \frac{1}{0.51} ight) = 4$$\frac{v}{2} \left( \frac{0.51 - 0.5}{0.5 	imes 0.51} ight) = 4$$\frac{v}{2} \left( \frac{0.01}{0.255} ight) = 4$$v \left( \frac{0.01}{0.51} ight) = 4$$v = \frac{4 	imes 0.51}{0.01} = 4 	imes 51 = 204 \,ms^{-1}$.

Two open organ pipes of lengths $50 \,cm$ and $51 \,cm$ are totally immersed in a medium. They are found to give $40$ beats in $10 \,s$ when each is sounding at its fundamental note. The speed of sound in this medium is (in $\,ms^{-1}$)

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