Under constant tension and constant mass per unit length, the note produced by a plucked string is 500Hz when the length of the string is 0.90m. At what length is the frequency 150Hz?
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Correct Answer: Option A
Explanation:
F = \(\frac{1}{2L}\)\(\sqrt\frac{T}{m}\)
F1L1 = F2L2
500 x 0.9 = 150 x L2
therefore, L2 = 3m
F = \(\frac{1}{2L}\)\(\sqrt\frac{T}{m}\)
F1L1 = F2L2
500 x 0.9 = 150 x L2
therefore, L2 = 3m