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(D) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_{o} = \frac{v}{2L}$ and that of a closed organ pipe is

Vedclass · Accepted Answer

$7$

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(D) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_{o} = \frac{v}{2L}$ and that of a closed organ pipe is $f_{c} = \frac{v}{4L}$.Given that $f_{o} - f_{c} = 2 	ext{ Hz}$.Substituting the expressions,$\frac{v}{2L} - \frac{v}{4L} = 2 \Rightarrow \frac{v}{4L} = 2 	ext{ Hz}$.Thus,$f_{c} = 2 	ext{ Hz}$ and $f_{o} = 2f_{c} = 4 	ext{ Hz}$.Now,the length of the open pipe is halved $(L_{o}' = L/2)$,so its new frequency is $f_{o}' = \frac{v}{2(L/2)} = \frac{v}{L} = 2f_{o} = 2 	imes 4 = 8 	ext{ Hz}$.The length of the closed pipe is doubled $(L_{c}' = 2L)$,so its new frequency is $f_{c}' = \frac{v}{4(2L)} = \frac{1}{2} \left(\frac{v}{4L}ight) = \frac{1}{2} f_{c} = \frac{1}{2} 	imes 2 = 1 	ext{ Hz}$.The number of beats produced per second is $|f_{o}' - f_{c}'| = |8 - 1| = 7 	ext{ Hz}$.

Similar Questions

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