91Ó°ÊÓ

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

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:

$\mathrm{λ}=\frac{v}{f}$

v= 344 m/s, the speed of sound in the air.

f = 150 Hz, the frequency of the given sound wave.

$$\begin{gathered} \mathrm{λ}=\frac{344m/s}{150Hz} \\ \mathrm{λ}=2.3m \end{gathered}$$

Hence

$$\begin{gathered} k=\frac{2\mathrm{π}}{\mathrm{λ}} \\ =2.74m^{-1} \end{gathered}$$

Substitute Into (1) for k, Band A to determine P_max

role="math" localid="1664346402579" $$\begin{gathered} P_{max}=\left(1.42×{10}^{5}Pa\right)×\left(2.74m^{-1}\right)×\left(0.02×{10}^{-3}m\right) \\ {P}_{max}=7.78Pa \end{gathered}$$

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:

$$\begin{gathered} \mathrm{λ}=\frac{344m/s}{1500Hz} \\ \mathrm{λ}=0.23m \end{gathered}$$

Hence

$$\begin{gathered} k=\frac{2\mathrm{π}}{\mathrm{λ}} \\ =27.3m^{-1} \end{gathered}$$

Substitute into (1) for k, B and A to determine Pmax

$$\begin{gathered} P_{max}=\left(1.42×{10}^{5}Pa\right)×\left(27.3m^{-1}\right)×\left(0.02×{10}^{-3}m\right) \\ {P}_{max}=77.8Pa \end{gathered}$$

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:

$$\begin{gathered} \mathrm{λ}=\frac{344m/s}{15000Hz} \\ \mathrm{λ}=0.023m \end{gathered}$$

$$\begin{gathered} k=\frac{2\mathrm{π}}{\mathrm{λ}} \\ =273m^{-1} \end{gathered}$$

Substitute Into (1) for A, B and A to determine Pmax

$$\begin{gathered} P_{max}=\left(1.42×{10}^{5}Pa\right)×\left(273m^{-1}\right)×\left(0.02×{10}^{-3}m\right) \\ {P}_{max}=778Pa \end{gathered}$$

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.