Two loudspeakers are 1 meter apart. A person stands 12 m directly in front of one. What is the lowest f which destructive and constructive interference will be heardPlease explain and solve

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that a constructive interference happens when the phase difference yields `(2npi), ` while the destructive interference happens when the phase difference yields `(2n+1)pi` .

Hence, the condition for constructive interference is `Delta r= (2n)lambda/2` , while the condition for destructive interference is `Delta r = (2n+1)lambda/2.`

`Delta r`  denotes the path difference.

The problem provides the length of path between the loudspeakers and the length of path from one loudspeaker to the person. Since the person stands directly in front of one loudspeaker, the triangle formed by locations of loudpeakers and person is a right triangle. The path from person to one loudspeaker denotes a leg and the path from person to the other loudspeaker denotes the hypotenuse.

You should come up with the notations: `r_1`  = 1m -path between -2loudspeakers, `r_2 ` = 12m - path between one loudspeaker and person, `r_3`  - path between the other loudspeaker and person.

Using the Pythagorean theorem yields:

`r_3 = sqrt(r_1^2 + r_2^2)`

`` `r_3 = sqrt(1 + 144) =gt r_3 = sqrt145 = 12.04`

You need to find the path difference such that:

`Delta r = r_3 - r_2 = 12.04 - 12 = 0.04`

The lowest frequence happens if `delta r = lambda/2, => 0.04 = lambda/2 =gt lambda = 0.08` .

You need to remember the frequency formula such that:

`f = v/lambda`

Considering v the speed of sound in the air yields:

`f = 343/0.08 = 4287.5 Hz`

Hence, evaluating the lowest frequency yields `f = 4287.5 Hz.`

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