# Two small loudspeakers emit sound waves of the same amplitude and frequency in phase with each...

## Question:

Two small loudspeakers emit sound waves of the same amplitude and frequency in phase with each other. The waves produce an interference pattern such that the distance from the right bisector to a point P on the third nodal line is x3. The wavelength of the sound is increased by a factor of 4. What is the new distance?

1) 1/4 x3

2) x3

3) 4 x3

4) 8 x3

## Interference

Wave is the disturbance in a medium that propagates and carries energy with itself (except for light which propagates even in a vacuum). Interference of two waves is the phenomenon observed when two waves of the same frequency superpose in a region resulting in a wave of higher or lower amplitude. Two waves are said to interfere constructively at a location when crest of one wave falls on the crest of another. Similarly, destructive interference is observed when the crest of one wave falls on the trough of another.

Given:

• Distance from right bisector to the point of the third nodal line is: {eq}x_3 {/eq}

For the waves to interfere destructively (to form a node), one of the waves must travel an integral multiple of distance equal to {eq}\dfrac{\lambda}{2} {/eq}

As the distance from right bisector to the point {eq}P {/eq} is {eq}x_3 {/eq}, one wave has traveled an additional distance equal to {eq}2 x_3 {/eq}. Therefore, for the third nodal line:

{eq}2 \times x_3 \ = \ 3 \times \dfrac{\lambda}{2} {/eq}

Let {eq}x_{3'} {/eq} be the distance to the third nodal line when {eq}\lambda' \ = \ 4 \times \lambda {/eq}. Then,

{eq}2 \times x_{3'} \ = \ 3 \times \dfrac{\lambda'}{2} \ = \ 4 \times (2 \ x_3) \ \\ \implies \ x_{3'} \ = \ 4 \times x_3 {/eq} 