A piano wire 0.5m long has a total mass of 0.01kg and is stretched with a tension of 800N. Calculate the frequency of the wire when it sounds its fundamental note
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Correct Answer: Option A
Explanation:
F = \(\frac{1}{2L}\)\(\sqrt\frac {T}{m}\)
N:B m is mass per length
= \(\frac {0.01}{0.5}\)
= 0.02kgm-1
F = \(\frac{1}{2\times 0.5}\)\(\sqrt\frac {800}{0.02}\)
= 200Hz
F = \(\frac{1}{2L}\)\(\sqrt\frac {T}{m}\)
N:B m is mass per length
= \(\frac {0.01}{0.5}\)
= 0.02kgm-1
F = \(\frac{1}{2\times 0.5}\)\(\sqrt\frac {800}{0.02}\)
= 200Hz