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In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the position of one worker. The details of practical systems are complex, but we can present a simple example that gives you the idea. Suppose a machine 5.5 m away from a worker emits a persistent 90 Hz hum. To cancel the sound at the worker's location with a speaker that exactly duplicates the machine's hum, how far from the worker should the speaker be placed? Assume a sound speed of 340 m/s.

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Answer:

7.39 m or 3.61 m

Explanation:

[tex]\lambda[/tex] = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

[tex]s_1-s_2=\frac{\lambda}{2}[/tex]

Velocity of wave

[tex]v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m[/tex]

[tex]s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m[/tex]

So, the location from the worker is 7.39 m or 3.61 m

Cricetus

The worker should place the speaker "7.39 m" or "3.61 m" far.

Frequency and velocity

According to the question,

Frequency, f = 90 Hz

Speed of sound, v = 340 m/s

We know the relation,

Velocity of wave, v = fλ

or,

Wavelength, λ = [tex]\frac{v}{f}[/tex]

By substituting the values,

                        = [tex]\frac{340}{90}[/tex]

                        = 3.78 m

Now,

We know that the path difference be:

→ s₁ - s₂ = [tex]\frac{\lambda}{2}[/tex]

         s₁ = s₂ [tex]\pm[/tex] [tex]\frac{\lambda}{2}[/tex]

             = 5.5 [tex]\pm[/tex] [tex]\frac{3.78}{2}[/tex]

             = 7.39 m or,

             = 3.61 m

Thus the response above is appropriate.

Find out more information about velocity here:

https://brainly.com/question/6504879

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