 2022-01-05

A well with vertical sides and water at the bottom resonates at 7.00 Hz and at no lower frequency. The air-filled portion of the well acts as a tube with one closed end (at the bottom) and one open end (at the top).The air in the well has a density of $1.10×k\frac{g}{{m}^{3}}$ and a bulk modulus of
$1.33×{10}^{5}Pa$
How far down in the well is the water surface? Annie Gonzalez

The water of the bottom resonates at 7 Hz
The air in the well has a density of $1.1k\frac{g}{{m}^{3}}$ a bulk modulus of $1.33×{10}^{5}Pa$
The frequency of water is given by
$f=\frac{v}{\lambda }$
$\lambda =4L$, L is the depth of the well from the water surface to the open end at the top.
$L=\frac{v}{4f}$
$v=\sqrt{\frac{\beta }{\rho }}$$\beta$ is the bulk modules, $\rho$ is the density of air in the well
$L=\frac{1}{4f}\sqrt{\frac{\beta }{\rho }}$
$L=\frac{1}{4×7Hz}×\sqrt{\frac{1.33×{10}^{5}pa}{1.1k\frac{g}{{m}^{3}}}}$ $=12.4m$ (answer) eninsala06

Both the top of the well and the top of the water are displacement nodes and anti-nodes, respectively. Exactly one-fourth of a wavelength may fit within the well's depth at the lowest resonance frequency.

If d is the depth and $\lambda$ is the wavelength, then $\lambda =4d$

The frequency is $f=\frac{v}{\lambda }=\frac{v}{4d}$
Where v is the speed of sound. The speed of sound is given by $v=\sqrt{\frac{\beta }{\rho }}$$\beta$ is the bulk modulus $\rho$ is the density of air in the well. Thus $f=\left(\frac{1}{4d}\right)\sqrt{\frac{\beta }{\rho }}$ and $d=\frac{1}{4f}\sqrt[\beta ]{\rho }$
$=\frac{1}{4\left(7.00Hz\right)}\sqrt{\frac{1.33×{10}^{5}Pa}{1.10k\frac{g}{{m}^{3}}}}$$=12.4m$ karton