Question

A tuning for vibrating at frequency $800\text{Hz}$ produces resonance in a resonance tube. The upper end is open, and the lower end is closed by the water surface, which can be varied. Successive resonance are observed at length $9.75\text{cm},31.25\text{cm}\text{and}52.75\text{cm}$. The speed $\left(\text{in m/s}\right)$ of sound in air from these data is _____.

Answer 1

For the tube open at one end, resonance frequencies are $\frac{nv}{4l}$ where $n$ is a positive odd integer.

Given that, the frequency of tuning fork, $f=800\text{Hz}$, successive length of tube in resonance with tuning fork is $l_1=9.75\text{cm},l_2=31.25\text{cm},l_3=52.75\text{cm}$

Now, as the same tuning fork produces successive resonance,

$f=\frac{nv}{4l_1}$

$\Rightarrow l_1=\frac{nv}{4f}$

$f=\frac{(n+2)v}{4l_2}$

$\Rightarrow l_2=\frac{(n+2)v}{4f}$

$f=\frac{(n+4)v}{4l_3}$

$\Rightarrow l_3=\frac{(n+4)v}{4f}$

$\Rightarrow l_3-l_2=l_2-l_1=\frac{2v}{4f}=\frac{v}{2f}$

With the given data,

$l_3-l_2=52.75-31.25=21.50\text{cm}$

$\Rightarrow l_3-l_2=0.2150\text{m}$

$\Rightarrow 0.2150=\frac{v}{2\times 800}$

$\Rightarrow v=344{\text{ms}}^{-1}$

Hence, $344$ is the correct answer..