The fourth overtone of a closed pipes is 900Hz, its fundamental frequency is
The correct answer is: B
Explanation
For a closed pipe, only odd harmonics are present. first overtone frequency is 3f\(_o\), 2\(^{nd}\) overtone frequency is 5f\(_o\) etc.
4\(^{th}\) overtone frequency = 9f\(_o\)
900 = 9f\(_o\)
f\(_o\) = \(\frac{900}{9}\) = 100Hz
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