Question

An organ pipe open at both ends sounds in unison with a tuning fork at ${20}^{\circ }C$. When the fork and the pipe are sounded together at ${30}^{\circ }C$. 5 beats per second are heard. If the frequency of the fork is approximately $37\beta$, find the value of $\beta$. Assume that frequency of fork is not affected by the temperature change. ___

Answer 1

Let the frequency of tuning fork is t Hz. Let the frequency of the note from the open organ pipe at ${20}^{\circ }C$ and ${30}^{\circ }C$ be $t_1$ and $t_2$ respectively. At ${20}^{\circ }C$, unison happens, So,
$t=t_1$
At ${30}^{\circ }C$, 5 beats per second are heard. So,
$t_2-t=5$
$t-\sqrt{T}=\frac{t_1}{t_2}-\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{293}{303}}=0.983$
$\therefore \frac{t}{5+t}=0.983\Rightarrow t=\left(0.983\right)5+\left(0.983\right)t\Rightarrow t=289Hz$
$37\beta =289\Rightarrow \beta =7.8=8\left(app\right)$