Question

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0×10^−5 Pa.

Part A) Calculate the intensity of this sound wave at 20 ∘C.

Part B) Calculate the sound intensity level of this sound wave at 20 ∘C.

Part C) Calculate the displacement amplitude of this sound wave at 20 ∘C.

Answer 1

The intensity of this sound wave at \(20^{\circ} \mathrm{C}\) is,

$$ \begin{array}{l} l=\frac{P^{2} v}{2 B} \\ I=\frac{\left(6.0 \times 10^{-5} \mathrm{~Pa}\right)^{2}(344 \mathrm{~m} / \mathrm{s})}{2\left(1.42 \times 10^{5} \mathrm{~Pa}\right)} \\ I=4.36 \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2} \end{array} $$

(b) The sound intensity level at \(20^{\circ} \mathrm{C}\) is,

$$ \begin{aligned} \beta &=(10 \mathrm{~dB}) \log \frac{4.36 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}{10^{-12}} \\ &=6.4 \mathrm{~dB} \end{aligned} $$

\((c)\)

The displacement amplitude of this sound wave at \(20^{\circ} \mathrm{C}\) is,

$$ \begin{aligned} A &=\frac{P_{v}}{2 \pi f B} \\ &=\frac{\left(6.0 \times 10^{-5} \mathrm{~Pa}\right)(344 \mathrm{~m} / \mathrm{s})}{2 \pi(400 \mathrm{~Hz})\left(1.42 \times 10^{5} \mathrm{~Pa}\right)} \\ &=5.78 \times 10^{-11} \mathrm{~m} \end{aligned} $$