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A loudspeaker at the origin emits a 110 Hz tone on a day when the speed of sound is 340 m/s. The phase difference between two points on the xxx-axis is 6.0 rad .
What is the distance between these two points? Express your answer with the appropriate units.

Respuesta :

The distance between the two points on a loudspeaker that initially emits a 110 Hz tone with a sound speed of 340 m/s = 18.54 m.

Traveling and stationary waves

Traveling waves are waves that have a fixed amplitude. Meanwhile, a stationary wave is a wave whose amplitude changes at every point.

The phase difference is the difference in the phases of the waves or stages of the waves. Mathematically, the phase difference is formulated as follows:

Δφ = Δx/λ

We have,

The frequency =  (110 Hz) ⇒ f

The speed of sound  = 340 m/s ⇒ v

The phase difference between two points =  6.0 rad ⇒ Δφ

Determine the wavelength first,

λ = v/f

= 340/110

= 3.09 m

So, the distance between these two points:

Δφ = Δx/λ

(6.0) = Δx/3.09

Δx = 18.54 m

Learn more about wavelength here: https://brainly.com/question/29800252

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