## Related questions with answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A loud factory machine produces sound having a displacement amplitude of $1.00 \mu \mathrm{m}$, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is $1.42 \times 10^{5} \mathrm{Pa}$. What is the highest frequency sound to which this machine can be adjusted without exceeding the prescribed limit? Is this frequency audible to the workers?

Solution

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

$\begin{align} P_{\text{max}}=BkA \end{align}$

Where the bulk modulus of the air is $B=1.42 \times 10^{5}$ Pa and the displacement amplitude of the waves produced by the machine is $1 \mathrm{~ \mu m}$.

Using (1) we can calculate $k$ then we can use $k$ to determine the wavelength $\lambda$ of the wave, and remember that $\lambda=2\pi/k$.

So, substitute into (1) with 10 Pa for $P_{\text{max}}$, $(1.42 \times 10^{5}$ Pa) for $B$ and $1\times 10^{-6} \mathrm{~ m}$ for $A$

$10 \mathrm{~ Pa}=( 1.42 \times 10^{5} \mathrm{~ Pa} ) \times k\times (1\times 10^{-6} \mathrm{~ m})$

$k=\frac{10 \mathrm{~ Pa}}{( 1.42 \times 10^{5} \mathrm{~ Pa} ) \times (1\times 10^{-6} \mathrm{~ m})}$

$k=70.4 \mathrm{~ m^{-1}}$

We can use the following relation to calculate the wavelength

$\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{70.4 \mathrm{~ m^{-1}}}$

$\lambda=0.089 \mathrm{~ m}$

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

$\begin{align} f=\frac{v}{\lambda} \end{align}$

$f=\frac{344 \mathrm{~ m/s}}{0.089 \mathrm{~ m}}$

$\boxed{f=3.86 \times 10^{3}\mathrm{~ Hz}}$

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

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