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  4. Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is

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Q. Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is 360 m/s, the shortest distance between two antinodes is

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Solution:

Shortest distance between two antinodes is equal to the half of wavelength.
Here, $v = 360 m s^{-1} , \upsilon = 200 \, Hz$
$ \therefore \, \lambda = \frac{v}{\upsilon} = \frac{360}{200} = 1.8 \, m $
Hence $ \frac{\lambda }{2} = \frac{1.8}{2} = 0.9 \, m $