Question

When you “crack” a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called “cavitation,” produces a sound pulse, the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside. If the pulse has a sound level of $\mathbf{\text{62dB}}$at your ear, estimate the rate at which energy is produced by the cavitation?

Short answer

The rate at which energy is produced by the cavitation is $2\text{\hspace{0.17em}μW}$.

Step-by-step solution

Given

1. Sound level (in dB):$\mathrm{\beta }=6\text{2\hspace{0.17em}dB}$
2. Distance from ear to knuckle cavity,$\mathrm{r}=0.\text{24\hspace{0.17em}m}$

Determining the concept

Find the intensity using the formula for sound level in terms of intensity and threshold intensity. Then, using the formula for power, find the energy rate.

The sound level (in dB) is given as-

1. $\mathrm{\beta }=10\log \left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)$
   

The power is related to intensity as -

ii.$\mathrm{P}=\frac{\mathrm{I}}{\mathrm{A}}$

where P is power;I, ${\mathrm{I}}_0$are intensities and A is area.

Determining the rate at which energy is produced by the cavitation

Let, β be the sound level and I be the intensity of the sound,

$\mathrm{\beta }=10\log \left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)$

Here,${\mathrm{I}}_0={10}^{-12}{\text{\hspace{0.17em}W/m}}^{\text{2}}$

Then,

$\begin{array}{c}\mathrm{I}={\mathrm{I}}_0\left({10}^{0.1\mathrm{\beta }}\right)\\ =({10}^{-12}\text{W}/{\text{m}}^{\text{2}})({10}^{0.1\left(62\text{\hspace{0.17em}dB}\right)})\\ =1.58\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Since, distance from knuckle to ear is,

$\mathrm{r}=0.24\text{m}$

So,

$\begin{array}{c}\text{Power}=\frac{\text{Intensity}}{\text{Area}}\\ =\frac{1.58\times {10}^{-6}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}}{4\left(3.14\right){\left(0.24\text{\hspace{0.17em}m}\right)}^2}\\ =2\text{μW}\end{array}$

Hence, the rate at which energy is produced by the cavitation is $2\text{μW}$.