Organ Pipe Overtones – Rankers Physics

Standing Wave in String and Organ Pipe: Practice Problem & Solution

The \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{rd}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
9 : 8
7 : 9
8 : 9
9 : 7

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Standing Wave in String and Organ Pipe. Understanding the underlying formula is key to arriving at the correct answer below:

For a closed pipe, \(f_c = \frac{9v}{4L_c}\). For an open pipe, \(f_o = \frac{4v}{2L_o}\). Since \(f_c = f_o\), we have \(\frac{9v}{4L_c} = \frac{4v}{2L_o} ⇒\frac{L_c}{L_o} = \frac{9}{8}\).

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