In a sound wave in air, the adjacent rarefactions and compressions are separated by a distance of 17cm. If the velocity of the sound wave is 340ms-1. Determine the frequency
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Correct Answer: Option C
Explanation:
\(\lambda\) = 2L
2 x 17 = 34cm
F = \(\frac{V}{\lambda}\)
= \(\frac{340}{34 \times 10^{-2}}\)
= 1000Hz
\(\lambda\) = 2L
2 x 17 = 34cm
F = \(\frac{V}{\lambda}\)
= \(\frac{340}{34 \times 10^{-2}}\)
= 1000Hz