Standing Wave in String and Organ Pipe: Practice Problem & Solution
The fundamental frequency of a closed organ pipe of length $20 \text{ cm}$ is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is: (2015)
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:
Fundamental of closed = $v / (4 \times 20)$. 2nd overtone of open = $3v / 2L$. So, $v / 80 = 3v / 2L \Rightarrow 2L = 240 \Rightarrow L = 120 \text{ cm}$.
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