Question

An organ pipe is sounded with a tuning fork of frequency $256$ Hz. When the air in the pipe is at a temperature of ${15}^o$C, $23$ beats are heard in $10$ sec. When the tuning fork is loaded with a small piece of wax, the beat frequency is found to decrease. What change of temperature of the air in the pipe is necessary to bring the pipe and the unloaded fork into unison?

Answer 1

Let frequency of the tuning fork $=f_1=256$Hz.
$F_b=|f_t-f_{pipe}|=\frac{23}{10}$
When the tuning fork is loaded, frequency is decreased. As beat frequency is decreased, $f_t>f_p$
$\Rightarrow f_t=f_{pipe}+2.3$ Putting, $f_t=256$ Hz
We obtain $f_{pipe}=253.7$ Hz
Let the temperature should decrease by $\theta$, the frequency of the pipe increases from $253.7$ Hz to $256$ Hz
$\Rightarrow \frac{256}{253.7}\cong 1+\frac{\theta }{576}$
$\Rightarrow \frac{\theta }{576}=\frac{256}{253.7}-1=\frac{2.3}{253.7}$
$\Rightarrow \theta =\left(\frac{23}{2537}\right)\left(576\right)={5.2}^o$C