Adela Brown

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 $1.10\times k\frac{g}{{m}^{3}}$ and a bulk modulus of

$1.33\times {10}^{5}Pa$

How far down in the well is the water surface?

How far down in the well is the water surface?

Annie Gonzalez

Beginner2022-01-06Added 41 answers

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 $1.33\times {10}^{5}Pa$

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

$L=\frac{1}{4f}\sqrt{\frac{\beta}{\rho}}$

$L=\frac{1}{4\times 7Hz}\times \sqrt{\frac{1.33\times {10}^{5}pa}{1.1k\frac{g}{{m}^{3}}}}$ $=12.4m$ (answer)

eninsala06

Beginner2022-01-07Added 37 answers

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\times {10}^{5}Pa}{1.10k\frac{g}{{m}^{3}}}}$$=12.4m$

karton

Expert2022-01-11Added 613 answers

One might imagine the well as an open organ pipe. The formula for its fundamental frequency is

$n=\frac{v}{41}\text{}and\text{}v=\sqrt{\frac{\beta}{\rho}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{\sqrt{\frac{\beta}{\rho}}}{41}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{\sqrt{\frac{\beta}{\rho}}}{4n}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{\frac{1.33\times {10}^{5}}{1.1}}}{4\times 7}\phantom{\rule{0ex}{0ex}}=12.4meter.$

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