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 \(3^{\text{th}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
Solution Explained:
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:
The frequency of the \(4^{\text{th}}\) overtone (9th harmonic) of a closed pipe is \(f_c = \frac{9v}{4L_c}\). The frequency of the \(3^{\text{rd}}\) overtone (4th harmonic) of an open pipe is \(f_o = \frac{4v}{2L_o} = \frac{2v}{L_o}\). Equating the two, \(\frac{9v}{4L_c} = \frac{2v}{L_o} ⇒ \frac{L_c}{L_o} = \frac{9}{8}\).
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