A resonance occurs with a tuning fork and an air column of size `12 cm`. The next higher resonance occurs with an air column of `38 cm`. What is the frequency of the tuning fork ? Assume that the speed of sound is `312 m//s`.

Options

A. `500 Hz`
B. `550 Hz`
C. `600 Hz`
D. `650 Hz`

Answer

C

Explanation

Since the standing wave mode has a displacement antinode at the opening , there is a displacement node at the water - air interface . By increasing the height of the air column , to go from one harmonic to the next , an additional length equal to `1//2` wavel,ength is requires . Hence
`(lambda)/(2) = (0.38 - 0.12)m rArr lambda = 0.52 m`
Finally , from ` v = f lambda` , we find that ` f = v//lambda = 312//0.52 = 600 Hz`. If one checks , this problem deals with the `i st and 3 rd `harmonics.