Question

A column of air and a tuning fork produces $4$ beats per second. When sounding together, tuning fork gives the lower note. The temperature of air is $16^{\circ} C$. When the temperature falls to $10^{\circ} C$, the two produce $3$ beats per second. What is the frequency (in $Hz$ ) of tuning fork? (Take $\sqrt{283}=16.82$ )

Answer 1

$v=\sqrt{\frac{r R T}{M}}$ ...(i) and
$f=\frac{v}{\lambda}$ ...(ii)
From (i) and (ii)
$\Rightarrow f \propto \sqrt{ T }$ ...(iii)
Let the frequency of tuning fork be $f$.
Thus, frequency of air column at $16^{\circ} C (289\, K )$
$=f+4$
and frequency of air column at $10^{\circ} C (283\, K )$
$=f+3$
From (iii)
$\Rightarrow \frac{f+4}{f+3}=\sqrt{\frac{289}{283}}$
$=\frac{17}{\sqrt{283}}=\frac{17}{16.82}$
$\Rightarrow \frac{f+4}{f+3}=1.011$
$\Rightarrow f+4=1.011 f+3 \times 1.011$
$\Rightarrow(1.011-1) f=4-3.033$
$\Rightarrow f=\frac{0.967}{0.011}=87.91\, Hz$