Question

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog’s mouth but by its eardrums, which lie on the surface of the head. And, surprisingly, the sound has nothing to do with the frog’s inflated throat. If the emitted sound has a frequency of $\mathbf{\text{260Hz}}$and a sound level of$\mathbf{\text{85dB}}$(near the eardrum), what is the amplitude of the eardrum’s oscillation? The air density is${\mathbf{\text{1.21kg/m}}}^{\mathbf{\text{3}}}$.

Short answer

The eardrum’s oscillations has the amplitude of $0.\text{76\hspace{0.17em}\hspace{0.17em}μm}$.

Step-by-step solution

Given

1. Frequency: f = 260 Hz
2. Air density:$\mathrm{\rho }=1.21{\text{\hspace{0.17em}kg/m}}^{\text{2}}$
3. Level of intensity:$\beta =85\text{\hspace{0.17em}dB}$

Determining the concept

Find the sound intensity using the formula for the sound level in terms of intensity. Write the amplitude in terms of intensity, density, frequency and speed of sound.

Formulae are as follow:

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

$\mathrm{I}=\frac{1}{2}\mathrm{\rho v}{\left(2\mathrm{\pi f}\right)}^2{\mathrm{A}}^2$

where, I,${\mathrm{I}}_0$ are intensities, f is frequency, A is area,$\mathrm{\rho }$ is density and v is density.

Determining the amplitude of the eardrum’s oscillation 

From the formula for sound level,

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

$\begin{array}{c}\mathrm{I}={\mathrm{I}}_0{10}^{0.1\mathrm{\beta }}\\ =\left({10}^{-12}{\text{\hspace{0.17em}W/m}}^2\right)\left({10}^{8.5\text{\hspace{0.17em}dB}}\right)\\ =3.162\times {10}^{-4}{\text{W/m}}^{\text{2}}\end{array}$

Since, intensity of sound is defined as,

$\mathrm{I}=\frac{1}{2}\mathrm{\rho v}{\left(2\mathrm{\pi f}\right)}^2{\mathrm{A}}^2$

$\begin{array}{c}\mathrm{A}=\sqrt{\frac{2\mathrm{I}}{\mathrm{\rho v}{\left(2\mathrm{\pi f}\right)}^2}}\\ =\sqrt{\frac{2\times (3.162\times {10}^{-4}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}})}{(\text{1.21\hspace{0.17em}\hspace{0.17em}kg}/{\text{m}}^{\text{3}})(340\text{\hspace{0.17em}m}/\text{s}){(2\times 3.14\times 260\text{\hspace{0.17em}Hz})}^2}}\\ =0.\text{76\hspace{0.17em}\hspace{0.17em}μm}\end{array}$

Hence, the oscillations has the amplitude equal to$0.\text{76\hspace{0.17em}\hspace{0.17em}μm}$ .