JAMB - Physics (1989 - No. 15)
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
10Hz
20Hz
1000Hz
5780Hz
Explicaţie
\(\lambda\) = 2L
2 x 17 = 34cm
F = \(\frac{V}{\lambda}\)
= \(\frac{340}{34 \times 10^{-2}}\)
= 1000Hz
2 x 17 = 34cm
F = \(\frac{V}{\lambda}\)
= \(\frac{340}{34 \times 10^{-2}}\)
= 1000Hz