91Ó°ÊÓ

In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.

Step by step solution

01

Finding the maximum speed of an element of air in the sound wave

We are given the hint to use Eq. (10-21), which is: $$ I = \frac{1}{2} \rho v^2 A_{max} \omega^2 $$ Where \(I\) is the intensity of the sound wave, \(\rho\) is the density of air, \(v\) is the speed of sound, \(A_{max}\) is the maximum displacement amplitude of the air particles, and \(\omega\) is the angular frequency. To find the maximum speed of an element of air in the sound wave \(v_{max}\), we can use the equation: $$ v_{max} = A_{max} \omega $$ First, rearrange Eq. (10-21) to solve for \(A_{max} \omega\): $$ A_{max} \omega = \sqrt{\frac{2I}{\rho v^2}} $$ Then substitute this expression into the equation for \(v_{max}\): $$ v_{max} = \sqrt{\frac{2I}{\rho v^2}} $$ Now, we can plug in the given values for \(I\), \(\rho\), and \(v\) to find \(v_{max}\): $$ v_{max} = \sqrt{\frac{2 \times 1.0 \times 10^{-12}W/m^2}{1.3 kg/m^3 \times (340 m/s)^2}} \approx 1.05 \times 10^{-5} m/s $$

02

Finding the average kinetic energy of the eardrum

We are given that the eardrum has mass \(m \approx 0.1 g\) and vibrates with displacement \(s_0\) at angular frequency \(\omega\). The maximum speed of the eardrum is equal to the maximum speed of an air element, which we found in Step 1 to be approximately \(1.05 \times 10^{-5} m/s\). The average kinetic energy of the eardrum is given by: $$ KE_{avg} = \frac{1}{2} m v_{max}^2 $$ Converting the eardrum mass to kilograms, we have \(m = 0.1 g = 10^{-4} kg\). Now, plug in the values for \(m\) and \(v_{max}\) to find the average kinetic energy: $$ KE_{avg} = \frac{1}{2} \times 10^{-4} kg \times (1.05 \times 10^{-5} m/s)^2 \approx 5.5 \times 10^{-21} J $$

03

Comparison and discussion

We are given that the average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} J\). Comparing our calculated average kinetic energy of the eardrum in the presence of the sound wave (\(5.5 \times 10^{-21} J\)) to this value, we can see that the energy in the presence of the sound wave is about one order of magnitude smaller. This implies that the human ear is incredibly sensitive, being able to detect sound waves with very small energies that barely cause the eardrum to vibrate. The fact that our auditory system can distinguish sound waves with such small energy differences is an amazing feat of natural engineering.

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