contestada

Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the two speakers and on the line separating them, thus creating a constructive interference at the listener's ear. What minimum distance would one of the speakers be moved back away from the listener to produce destructive interference at the listener's ear? (The speed of sound = 340 m/s.)

Respuesta :

cjmejiab

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

[tex]\lambda = \frac{v}{f}[/tex]

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

[tex]\lambda = \frac{340}{500}[/tex]

\lambda = 0.68m

The distance to move one speaker is half this

[tex]\lambda/2 = 0.34m[/tex]

Therefore the minimum distance will be 0.34m

The minimum distance that one of the speakers should be moved back away from the listener to produce destructive interference at the listener's ear is; Δ = 0.34 m

The formula for wavelength here as it relates to speed and frequency is given as;

λ = v/f

Where;

λ is wavelength

v is speed

f is frequency

We are given;

Frequency; f = 500 Hz

Speed of sound; v = 340 m/s

Thus;

λ = 340/500

λ = 0.68 m

Now, we are told that the line separating them creates a constructive interference at the listeners ear. Thus;

To calculate the minimum distance would one of the speakers be moved back away from the listener to produce destructive interference at the listener's ear we will use the formula;

formula for destructive path length is;

Δ = (m + ½)λ

And m here is zero

Thus;

Δ = ½λ

Δ = 0.68/2

Δ = 0.34 m

Read more at; https://brainly.com/question/14718577

Otras preguntas