Stationary waves of frequency 200 are formed in air. If velovity of the wave is 360 m/s the shortest distance between antinodes will be :

Options

A. 1.8 m
B. 3.6 m
C. 2.7 m
D. 0.9 m

Answer

D

Explanation

V = 200 Hz. V = 360 `ms^(-1)`
`lambda = (u)/(v) = (360)/(200) = 1.8 `m
distance bet. Two conseutive antinodes
`lambda// 2 = (18)/(2) = 0.9` m.
Hence correct choice is (d).