91Ó°ÊÓ

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

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

Where the bulk modulus of the air is$B=1.42×{10}^{5}Pa$ and the displacement amplitude of the waves produced by the machine is 1 μ-m.

Use equation (1) to calculate and then use k to determine the wavelength of the wave,

$\mathrm{λ}=\frac{2\mathrm{π}}{\mathrm{λ}}$

Substitute into equation (1) with 10 Pa for P_max, 1.42 x 105 Pafor B and 1 x 10-6 m for A:
$$\begin{gathered} k=\frac{10Pa}{\left(1.42×{10}^{5}Pa\right)×\left(1×{10}^{-6}m\right)} \\ k=70.4m^{-1} \end{gathered}$$

Use the following relation to calculate the wavelength:

$$\begin{gathered} \mathrm{λ}=\frac{2\mathrm{π}}{k} \\ =\frac{2\mathrm{π}}{70.4m^{-1}} \\ \mathrm{λ}=0.089m \end{gathered}$$

Finally, the relation between the wavelength and the frequency of a sound wave is given by the following equation:

$$\begin{gathered} f=\frac{v}{\mathrm{λ}} \\ f=\frac{344m/s}{0.089m} \\ f=3.86×{10}^{3}Hz \end{gathered}$$

Since is in the range of [20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.


Therefore, the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit is, f = 3.86 x 10³ Hz. Since f is in the range of [20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible to the workers.