Question

A well with vertical sides and water at the bottom resonates at 7.00Hzand at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of 1.10kg/m³and a bulk modulus of$\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathit{P}\mathit{a}$. How far down in the well is the water surface?

Short answer

The water surface is 12.4m down in the well.

Step-by-step solution

The given data

1. The density of the air, $\left(\rho \right)=1.10kg/m^3$
2. Resonant frequency$\left(f_r\right)=7.00Hz$
3. Bulk modulus$\left(\beta \right)=1.33\times {10}^{5}Pa.$

Understanding the concept of sound waves

We can find the sound velocity from bulk modulus and air density using the corresponding relation. By inserting it in the formula for the resonant frequency of pipe closed at one end for the lowest frequency, we can get how far down in the well the water surface is.

Formula:

The velocity of the sound wave in air,$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{\beta }}{\mathbf{\rho }}}$ …(1)

The resonant frequency of a wave, ${\mathit{f}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{nv}}{\mathbf{4}\mathbf{L}}$ …(2)

Calculation of the water surface level

Sound velocity is given by the formula from equation (1) and the given values as:

$\begin{array}{rcl}v& =& \sqrt{\frac{1.33\times {10}^{5}Pa}{1.10s}}\\ & =& 347.71m/s\end{array}$

The length of the wall using equation (2) is given as:

$L=\frac{1}{4\times f_r}v$

Substitute all the value in the above equation.

$\begin{array}{rcl}L& =& \frac{1}{4\times f_r}v\\ & =& \frac{1}{4\times 7Hz}\times 347.71m/s\\ & =& 12.418m\\ & \cong & 12.4m.\end{array}$

Hence, the level of the water in the well is 12.4m down.