A loud factory machine produces sound having a displacement amplitude in air of 1.00 μmμm, 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 PaPa. Under the conditions of this factory, the bulk modulus of air is 1.30×105 PaPa . The speed of sound in air is 344 m/s
What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?

f =3.4*10^3 hz

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

Explanation:

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

P_MAX = B*k*A                                              (1)  

Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement  amplitude of the waves produced by the machine is 1.00 μmμm.  

Using (1) we can calculate k then we can use k to determine the wavelength  A of the wave, and remember that λ = 2π/k.  

So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and  

1 x 10^-6 m for A  

10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)  

k = 62.5 m^-1

We can use the following relation to calculate the wavelength  

λ = 2π/k.  

λ = 0.100 m

Finally, the relation between the wavelength and the frequency of a sound  

wave is given by the following equation  

f = v/ λ

 =344/0.100 m

f =3.4*10^3 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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explanation: