Standing Wave in String and Organ Pipe: Practice Problem & Solution
The two nearest harmonics of an open organ pipe are 300 Hz and 450 Hz. If speed of sound in the pipe is 300 m/s, then length of the pipe is
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 an open organ pipe, successive harmonics differ by the fundamental frequency: \( f_1 = 450 - 300 = 150\text{ Hz} \). Using \( f_1 = \frac{v}{2L} \), we get \( 150 = \frac{300}{2L} \implies L = 1\text{ m} = 100\text{ cm} \).
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