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

Let the frequency of tuning fork be v. As `v=(upsilon)/(lambda)`, therefore, with decrease in temperature, `upsilon` decreases and hence v decreases. As number of beats `//` sec decreases with decrease in frequency of air column, therefore, frequency of air column is greater than the frequency of funing fork.
`:.` Frequency of air column at `51^(@)C=(V+4),`
and frequency of air column at `16^(@)C=(v+1)`
Now, `(v+4)/(v+1)=(upsilon_(51))/(upsilon_(16))=sqrt((273+51)/(273+16))=sqrt((324)/(289))=(18)/(17)`
`18v+18=17v+68`
`v=68-18=50Hz`