# Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of...

## Question:

Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 6.0 m, then forward by 4.0 m. Afterward, constructive interference is observed 1/4 and 3/4 of the distance between the speakers along the line that joins them.

What is the maximum possible wavelength of the sound waves?

## Constructive Interference:

Constructive interference takes place when two-in phase waves superimpose each other. In constructive interference, the individual maximum amplitude of each waves adds up producing a wave with a higher amplitude which resulted to the combination of the individual amplitude of the two waves.

First, using the Pythagorean Theorem, we can compute for the line distance between the speakers as,

{eq}d = \sqrt {{4.0 \ m)^2 + (6.0 \ m}^2} = 7.211102551 \ m {/eq}

As stated in the problem, the anti nodes in between the two speakers exist at {eq}\frac {1}{4} {/eq} and {eq}\frac {3}{4} {/eq}

The distance between the two consecutive anti nodes therefore is,

{eq}7.211102551 \ m * 0.75 - 7.211102551 \ m * 0.25 = 3.605551275 \ m {/eq}

Note that the wavelength in terms of the distance between anti nodes is given by,

{eq}\frac {\lambda}{2} = 3.605551275 \ m {/eq}

The wavelength therefore is,

{eq}\lambda = 2 * 3.605551275 \ m = 7.21110255 \ m = \boxed {7.2 \ m} {/eq}