For sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 x 108 m produces a pressure amplitude of 3.0 × 10-² Pa. Water at 20°C has a bulk modulus of 2.2 x 10⁹ Pa, and the speed of sound in water at this temperature is 1480 m/s. For 1000-Hz sound waves in 20°C water, what displacement amplitude is produced if the pressure amplitude is 3.0 x 102 Pa? Explain why your answer is much less than 1.2 x 10-8 m.

A sound wave with a frequency of 1000 Hz travels through the water with a speed of 1480 m/s. If the bulk modulus of the water is 2.2 x 10° Pa, determine the displacement amplitude that would be needed to produce a pressure amplitude of 3 × 10^-2 Pa.

The relation that describes the pressure amplitude for a sound wave is as follows:

$P_{max}=BkA$--(1)

In order to make use of equation (1) first calculate k:

$k=\frac{2\pi }{\lambda }$
So, the relation between the wavelength and the frequency of a sound wave is given by the following equation:

$\lambda =\frac{v}{f}$--(2)

v=1480 m/s, the speed of sound in the water.

f = 1000 Hz, the frequency of the given sound wave.

$$\begin{gathered} \lambda =\frac{1480m/s}{1000Hz} \\ \lambda =1.48m \end{gathered}$$

Therefore,
$$\begin{gathered} k=\frac{2\pi }{\lambda } \\ =4.25m^{-1} \end{gathered}$$

Now substitute the values of B, K, and P_max, into equation (1) to get A:

$$\begin{gathered} 3\times {10}^{-2}Pa=\left(2.2\times {10}^{9}Pa\right)\times \left(4.25m^{-1}\right) \\ A=\frac{3\times {10}^{-2}Pa}{\left(2.2\times {10}^{9}Pa\right)\times \left(4.25m^{-1}\right)} \\ A=3.2\times {10}^{-12}m \end{gathered}$$

A Is Inversely proportional to Bk, so, the reason for A being smaller than (1.2 x 10-8m) is that Bk for water is much larger than Bk for air.

Therefore, A=3.2 x 10-12 m is the displacement amplitude is produced if the pressure amplitude is 3.0 x 102 Pa. The reason for A being smaller than 1.2 x 10-8 Is that Bk for water is much larger than Bk for air.