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The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 \mathrm{~cm}. The length of the open organ pipe is __________\mathrm{cm}.

Option: 1

80


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

For open organ pipe ,
\mathrm{f_{n}= \frac{nv}{2l}\: \rightarrow (1)}

For closed organ pipe
\mathrm{f_{n}= \frac{\left ( 2n-1 \right )v}{4l}\: \rightarrow (2)}

For first overtone of an open organ pipe
\mathrm{n=2\left(2^{n d}\right. mode )}

\mathrm{f_{1}=\frac{2 \mathrm{~v}}{2 l_{1}} \rightarrow (3)}
Fundamental frequency of closed organ pipe, \mathrm{n=1}
\mathrm{f_{2} =\frac{v}{4 l_{2}} \rightarrow (4)}

\mathrm{f_{1} =f_{2} \quad \text { (Given) } }
\mathrm{\frac{2 v}{2 l_{1}} =\frac{v}{4 l_{2}} }
\mathrm{l_{1} =4 l_{2} }
     \mathrm{=4(20) }
     \mathrm{=80 \mathrm{~cm} }

The length of the open organ pipe is \mathrm{80 \mathrm{~cm} }.
 

Posted by

sudhir.kumar

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