Two waves of wavelength `99cm` and `100cm` both travelling with the velocity of `396ms^(-1)` are made to interfere. Calculate the number of beats produced `b` then per sec.

Explanation

`lamda_1 = 99 cm = 0.99 cm`
`lamda_2 = 100 cm = 1m `
Velocity of sound waves in a gas is ` v = 396 ms^(-1)`
Relation between wavelength and velocity `v = lamda f`
` therefore ` Frequency corresponding to ` lamda_1 , f_1 = (v)/(lamda_1) = (396)/(0.99) , f_1 = 400 Hz`
Frequency corresponding to `lamda_2 , f_2 = 396 Hz`
Number of beats per second ` |f_1 - f_2| = 400 -- 396`
n = 4 beats per sec