Q3E - University Physics with Modern Physics

Question

Consider a sound wave in air that has displacement amplitude 0.0200 mm. Calculate the pressure amplitude for frequencies of (a) 150 Hz; (b) 1500 Hz; (c) 15,000 Hz. In each case compare the result to the pain threshold, which is 30 Pa.

Step-by-Step Solution

Answer

a) Pmax= 7.78 Pa, smaller than the pain threshold. 
 
 b) Pmax P = 77.8 Pa, greater than the pain threshold. 
 
 c) Pmax = 778 Pa, greater than the pain threshold.

Step 1: Determination of the values given

A sound wave with a displacement amplitude of 0.02 mm traveling through the air is given.Calculate the pressure amplitude produced by such a wave if it has the following frequencies:
 
 (a) f = 150 Hz 
 
 (b) f = 1500 Hz 
 
 (c) f = 15000 Hz 

Step 2: Application of the formula of Sound and Hearing with given values

a) The relation that describes the pressure amplitude for a sound wave is as follows 
 
$P_{max}=BkA$ --(1) 
 
Where the bulk modulus of the air is B = 1.42 x 10⁵ Pa. Now, in order to make use of equation (1), first calculate k. So, the relation between the wavelength and the frequency of a sound wave given by the following equation:
 
$\lambda =\frac{v}{f}$
 
 v= 344 m/s, the speed of sound in the air. 
 
 f = 150 Hz, the frequency of the given sound wave. 
 

$$\begin{gathered} \lambda =\frac{344m/s}{150Hz} \\ \lambda =2.3m \end{gathered}$$
 
 Hence 
 
$$\begin{gathered} k=\frac{2\mathrm{\pi }}{\lambda } \\ =2.74m^{-1} \end{gathered}$$
 
Substitute Into (1) for k, Band A to determine P_max
 
$$\begin{gathered} P_{max}=\left(1.42\times {10}^5 Pa\right)\times \left(2.74m^{-1}\right)\times \left(0.02\times {10}^{-3} m\right) \\ {P}_{max}=7.78Pa \end{gathered}$$
 
Compare this value of P_max with the pain threshold which is 30 Pa, therefore it is smaller than the pain threshold. 
 
 b) For a wave with a frequency of 1500Hz:
 
$$\begin{gathered} \lambda =\frac{344m/s}{1500Hz} \\ \lambda =0.23m \end{gathered}$$ 

 Hence 
 
$$\begin{gathered} k=\frac{2\pi }{\lambda } \\ =27.3m^{-1} \end{gathered}$$
 
 Substitute into (1) for k, B and A to determine Pmax 
 
$$\begin{gathered} P_{max}=\left(1.42\times {10}^5 Pa\right)\times \left(27.3m^{-1}\right)\times \left(0.02\times {10}^{-3} m\right) \\ {P}_{max}=77.8Pa \end{gathered}$$
 
 Compare this value of Pmax with the pain threshold which is 30 Pa, therefore it is greater than the pain threshold. 
 
 c) For a wave with a frequency of 15000 Hz:
 
$$\begin{gathered} \lambda =\frac{344m/s}{15000Hz} \\ \lambda =0.023m \end{gathered}$$

 Hence,

$$\begin{gathered} k=\frac{2\pi }{\lambda } \\ =273m^{-1} \end{gathered}$$

 Substitute Into (1) for A, B and A to determine Pmax 
 
 $$\begin{gathered} P_{max}=\left(1.42\times {10}^5 Pa\right)\times \left(273m^{-1}\right)\times \left(0.02\times {10}^{-3} m\right) \\ {P}_{max}=778Pa \end{gathered}$$

 Compare this value of Pmax with the pain threshold which is 30 Pa, thereforeit is much greater than the pain threshold. 
 
 Therefore,

 a) Pmax= 7.78 Pa, smaller than the pain threshold. 
 
 b) Pmax P = 77.8 Pa, greater than the pain threshold. 
 
 c) Pmax = 778 Pa, greater than the pain threshold.