Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ metre long. The length of the open pipe will be

$L$

$2L$

$L/2$

$4L$

Step-by-Step Solution

Determine Frequency of Open Pipe: For an open organ pipe, the resonant frequencies are all harmonics, given by $f = n\frac{v}{2L_o}$ where $n = 1, 2, 3, \dots$ . The fundamental is $n=1$ , the first overtone is $n=2$ , and the second overtone is $n=3$ . So, the frequency of the second overtone is $f_o = \frac{3v}{2L_o}$ .
Determine Frequency of Closed Pipe: For a closed organ pipe, only odd harmonics are present, given by $f = (2n+1)\frac{v}{4L_c}$ where $n=0, 1, 2, \dots$ is the overtone number. For the first overtone ( $n=1$ ), the frequency is $f_c = \frac{3v}{4L_c}$ .
Equate Frequencies: The length of the closed pipe is given as $L_c = L$ . Since the two overtones have the same frequency ( $f_o = f_c$ ): $\frac{3v}{2L_o} = \frac{3v}{4L}$
Calculate Length: Solving for the length of the open pipe ( $L_o$ ): $\frac{1}{2L_o} = \frac{1}{4L} \implies 2L_o = 4L \implies L_o = 2L$

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