Sound waves from a tuning fork `A` reacha point `P` by two separate paths `ABP` and `ACP`. When `ACP` is greater than `ABP` by `11.5 cm`, there is silence at `p`. When the difference is `23 cm` the sound becomes loudest at `P` and `34.5 cm` there is silence again and so on. Calculate the minimum frequency of the fork if the velocity of sound is taken to be `331.2 m//s`.

Answer

1440 Hz

Explanation

Given that
`Delta_(1) = 11.5 cm = (2n + 1)lambda//2`
`Delta_(1) = 23 cm = (2n + 1)lambda`
`Delta_(1) = 34.5 cm = (2n + 1)(3lambda)/(2)`
If there is no maxima or minima between `Delta_(1), Delta_(2))` and `Delta_(3)` that means n = 0 for maximum wavelength or minimum frequency.
`(lambda)/(2) ` = 11.5 cm
`lambda` = 23 cm
`n = (v)/(lambda) = (331.2)/(0.23) ` = 1440 Hz