A column of air at `51^(@) C` and a tuning fork produce `4` beats per second when sounded together. As the temperature of the air column is decreased, the number of beats per second tends to decrease and when the temperature is `16^(@) C` the two produce `1` beat per second. Find the frequency of the tuning fork.

Explanation

The frequency of air column is given by
`n = (v)/(2l)`
Neglecting end correction, we have
`(n_(1))/(n_(2)) = (v_(1))/(v_(2)) = sqrt(((T_(1))/(T_(2))))`
`sqrt(((273 +51)/(273 + 16))) = sqrt(((324)/(289)))`
= 1.059
This shows that `n_(2) lt n_(1)`
As the number of beats with `n_(2)` is lessthan the number ofbeats with `n_(1)` hence the frequency ofair column must be greater than the frequency ofthe tuningfork. If n bethe frequency oftuning fork, then `n_(1) = n + 4` and` n_(2) = n + 2`
`(n+4)/(n+1) = 1.059`
= n = 49.8 Hz.