A column of air and a tuning fork produce 4 beats per second when sounded together. The tuning fork gives the lower note. The temperature of air is ${15}^{o}C$ . When the temperature falls to ${10}^{o}C$ , the two produce 3 beats per second. Find the frequency (in Hz) of the fork.

Let the frequency of the tuning fork be n Hz
Then frequency of air column at  ${15}^{o}C=n+4$
Frequency of air column at  ${10}^{o}C=n+3$
According to $v=n\lambda$ , we have
$v_{15}=(n+4)\lambda$ and  $v_{10}=(n+3)\lambda$
$\therefore \frac{v_{15}}{v_{10}}=\frac{n+4}{n+3}$
The speed of sound is directly proportional to the square root of the absolute temperature.
$\therefore \frac{v_{15}}{v_{10}}=\sqrt{\frac{15+273}{10+273}}=\sqrt{\frac{288}{283}}$
$\therefore \frac{n+4}{n+3}=\sqrt{\frac{288}{283}}={(1+\frac{5}{283})}^{1/2}\Rightarrow 1+\frac{1}{n+3}=1+1/2\times$
$\begin{array}{l}\frac{5}{283}=1+\frac{5}{566}\\ \Rightarrow \frac{1}{n+3}=\frac{5}{566}\\ \Rightarrow n+3=113\Rightarrow n=110Hz\end{array}$