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JAMB Physics 2024 Paper

Question : 38

Total: 125

The fourth overtone of a closed pipes is 900Hz, its fundamental frequency is

Solution:

For a closed pipe, only odd harmonics are present. first overtone frequency is 3f $_o$ , 2 $^{\text{nd}}$ overtone frequency is 5f $_o$ etc.

4 $^{\text{th}}$ overtone frequency = 9f $_o$

900 = 9f $_o$

f $_o$ = $\;\frac{900}{9}$ = 100Hz

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