Question

The air pressure variations in a sound wave cause the eardrum to vibrate. (a) For a given vibration amplitude, are the maximum velocity and acceleration of the eardrum greatest for high-frequency sounds or lowfrequency sounds? (b) Find the maximum velocity and acceleration of the eardrum for vibrations of amplitude \(1.0 \times 10^{-8} \mathrm{m}\) at a frequency of $20.0 \mathrm{Hz}\(. (c) Repeat (b) for the same amplitude but a frequency of \)20.0 \mathrm{kHz}$.

Short answer

Calculate the maximum velocity and acceleration for an eardrum with an amplitude of 1.0 x 10^-8 m, for sound waves with frequencies of 20 Hz and 20 kHz. Answer: The maximum velocity and acceleration of the eardrum are greater for high-frequency sounds. For a 20 Hz sound wave, the maximum velocity is 4.0π x 10^-7 m/s, and the maximum acceleration is 1.6π^2 x 10^-5 m/s². For a 20 kHz sound wave, the maximum velocity is 4.0π x 10^-4 m/s, and the maximum acceleration is 1.6π^2 x 10^-1 m/s².

Step-by-step solution

First derivative of displacement function (velocity)

To find the velocity of the eardrum, take the first derivative of the displacement function with respect to time: \(v(t) = \frac{dx(t)}{dt} = -Aω\sin(ωt)\)

Second derivative of displacement function (acceleration)

To find the acceleration of the eardrum, take the second derivative of the displacement function with respect to time: \(a(t) = \frac{d^2x(t)}{dt^2} = -Aω^2\cos(ωt)\)

Maximum velocity and acceleration dependencies on frequency

The maximum values of velocity and acceleration can be found by determining their absolute maximum values when taking into account the sine and cosine functions in these expressions: - The maximum velocity is given by \(|v(t)| = | - Aω\sin(ωt)| \leq Aω\). Thus, the maximum velocity depends on the angular frequency \(ω\), and so the maximum velocity is higher for high-frequency sounds. - The maximum acceleration is given by \(|a(t)| = | - Aω^2\cos(ωt)| \leq Aω^2\). Thus, the maximum acceleration also depends on the angular frequency \(ω\), and so the maximum acceleration is higher for high-frequency sounds.

Calculate maximum velocity and acceleration for 20 Hz sound wave

Given a vibration amplitude of \(1.0 \times 10^{-8} \mathrm{m}\) and a frequency of 20 Hz, we can calculate the maximum eardrum velocity and acceleration as follows: 1. Calculate the angular frequency: \(ω = 2πf = 2π(20 \mathrm{Hz}) = 40π\) rad/s 2. Calculate the maximum velocity: \(v_\text{max} = Aω = (1.0 \times 10^{-8} \mathrm{m})(40π \text{ rad/s}) = 4.0π \times 10^{-7}\) m/s 3. Calculate the maximum acceleration: \(a_\text{max} = Aω^2 = (1.0 \times 10^{-8} \mathrm{m})(40π \text{ rad/s})^2 = 1.6π^2 \times 10^{-5}\) m/s²

Calculate maximum velocity and acceleration for 20 kHz sound wave

Given the same vibration amplitude and a frequency of 20 kHz, we can calculate the maximum eardrum velocity and acceleration as follows: 1. Calculate the angular frequency: \(ω = 2πf = 2π(20,000 \mathrm{Hz}) = 40,000π\) rad/s 2. Calculate the maximum velocity: \(v_\text{max} = Aω = (1.0 \times 10^{-8} \mathrm{m})(40,000π \text{ rad/s}) = 4.0π \times 10^{-4}\) m/s 3. Calculate the maximum acceleration: \(a_\text{max} = Aω^2 = (1.0 \times 10^{-8} \mathrm{m})(40,000π \text{ rad/s})^2 = 1.6π^2 \times 10^{-1}\) m/s² So, for part (a), the greatest maximum velocity and acceleration are for high-frequency sounds. In part (b), the maximum velocity and acceleration for a 20 Hz sound wave are \(4.0π \times 10^{-7}\) m/s and \(1.6π^2 \times 10^{-5}\) m/s², respectively. In part (c), the maximum velocity and acceleration for a 20 kHz sound wave are \(4.0π \times 10^{-4}\) m/s and \(1.6π^2 \times 10^{-1}\) m/s², respectively.