Chapter 17: Q33P (page 508)

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 $260Hz$ and a sound level of $85dB$ (near the eardrum), what is the amplitude of the eardrum’s oscillation? The air density is ${1.21kg/m}^{3}$ .

Short Answer

Expert verified

The eardrum’s oscillations has the amplitude of $0. 76 μm$ .

Step by step solution

Given

Frequency: f = 260 Hz
Air density: $ρ = 1.21 {kg/m}^{2}$
Level of intensity: $β = 85 dB$

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:

$β = 10 \log (\frac{I}{I_{0}})$

$I = \frac{1}{2} ρv {(2 πf)}^{2} A^{2}$

where, I, $I_{0}$ are intensities, f is frequency, A is area, $ρ$ is density and v is density.

Determining the amplitude of the eardrum’s oscillation

From the formula for sound level,

$β = 10 \log (\frac{I}{I_{0}})$

$\begin{matrix} I = I_{0} 10^{0.1 β} \\ = (10^{- 12} {W/m}^{2}) (10^{8.5 dB}) \\ = 3.162 \times 10^{- 4} {W/m}^{2} \end{matrix}$

Since, intensity of sound is defined as,

$I = \frac{1}{2} ρv {(2 πf)}^{2} A^{2}$

$\begin{matrix} A = \sqrt{\frac{2 I}{ρv {(2 πf)}^{2}}} \\ = \sqrt{\frac{2 \times (3.162 \times 10^{- 4} W / m^{2})}{(1.21 kg / m^{3}) (340 m / s) {(2 \times 3.14 \times 260 Hz)}^{2}}} \\ = 0. 76 μm \end{matrix}$