The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe is $100\ Hz$ . The frequency of the third harmonic of the closed pipe is ..... $Hz$
Diffcult
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Let length of open organ pipe $=I_{0}$
Length of closed organ pipe $=\mathrm{I}_{\mathrm{c}}$
Also, $\mathrm{I}_{0}=2 \mathrm{I}_{\mathrm{C}}$
Fundamental frequency of the open pipe,
$v_{0}=100 \mathrm{\,Hz}$
Also, $\quad \mathrm{v}_{0}=\frac{\mathrm{v}}{2 \mathrm{l}_{0}}=100, \frac{\mathrm{v}}{\mathrm{l}_{0}}=200 ; \frac{\mathrm{v}}{2 \mathrm{l}_{\mathrm{C}}}=200, \frac{\mathrm{v}}{\mathrm{l}_{\mathrm{C}}}=400$
Fundamental frequency of closed organ pipe $v_{c}=\frac{v}{4 l_{c}}=\frac{400}{4}=100 \mathrm{\,Hz}$
So, the frequency of third harmonic of the closed organ pipe $=3 \mathrm{v}_{\mathrm{c}}=300 \mathrm{\,Hz}$
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