Consider two sound waves with wavelengths 5 m and 6 m . If these two waves propagate in a gas with velocity `330 ms^(-1)` . Calculate the number of beats per second .

Explanation

Given `lambda_(1) = 5m` and `lambda_(2) = 6` m
Velocity of sound waves in a gas is v = `330 ms^(-1)`
The relation between wavelength and velocity is `v = lambda f implies f = (v)/(lambda)`
The frequency corresponding to wavelength `lambda_(1)` is `f_(1) = (v)/(lambda_(1)) = (330)/(5) = 66` Hz
The frequency corresponding to wavelength `lambda_(1)` is `f_(2) = (v)/(lambda_(2)) = (330)/(6) = 55` Hz
The number of beats per second is `|f_(1) - f_(2) | = |66- 55| = 11` beats per sec