# The figure shows a loudspeaker A and point C, where a listener is positioned. |AC| = 5.00 m and...

## Question:

The figure shows a loudspeaker A and point C, where a listener is positioned. |AC| = 5.00 m and the angle {eq}\theta{/eq} = 44 degrees. A second loudspeaker B is located somewhere to the left of A. The speakers vibrate out of phase and are playing a 61.0 Hz tone. The speed of sound is 343 m/s. What is the third closest distance to speaker A that speaker B can be located, so that the listener hears no sound?

## Waves and Interference:

Wave is the propagation of a disturbance in a medium (or vacuum in the case of light). The wavelength of a wave ({eq}\lambda{/eq}) is the spatial separation between two adjacent locations where the wave pattern repeats itself. The frequency of a wave ({eq}\nu{/eq})is the number of wavelengths that traverse a particular location in one second. Speed of the wave is related to its wavelength and frequency as:

{eq}v \ = \ nu \times \lambda{/eq}

Interference is the phenomenon observed when two waves of the same frequency superpose to form a resultant wave of the same frequency and a different amplitude (usually). The waves are said to interfere constructively when the two waves meet up crest to crest at a given location. Destructive interference is observed when the crest of one wave falls on the trough of another.

## Answer and Explanation:

We are given:

- Distance from loudspeaker A to the point C: {eq}d_{AC} \ = \ 5 \ m {/eq}

- Angle made by the line AC with horizontal:{eq}\theta \ = \ 40^\circ \ deg{/eq}

- Frequency of the source: {eq}f \ = \ 60 \ Hz{/eq}

- Speed of sound: {eq}v \ = \ 343 \ ms^{-1} {/eq}

Let {eq}\lambda {/eq} be the wavelength of the source. Then:

{eq}lambda \ = \ \dfrac{v}{f} \ = \ \dfrac{343}{60} \ = \ 5.717 \ m {/eq}

Let {eq}d_{BC}{/eq} be the distance from point C to the loudspeaker B. As the two speakers are initially out of phase, one of the signals must travel an additional distance equal to an integral multiple of wavelength of the signal for them to interfere destructively at C i.e.:

{eq}d_{BC} \ - \ d_{AC} \ = \ n \times \lambda \ \\ \implies \ d_{BC} \ = \ d_{AC} \ + \ n \times lambda \ = \ 5 \ + \ 5.717 \times n {/eq}

The third closest distance corresponds to {eq}n \ = \ 3 {/eq}. Therefore:

{eq}d_{BC} \ = \ 5 \ + \ 3 \times 5.717 \ = \ 22.151 \ m {/eq}

The distance between the sources is:

{eq}d_{AB} \ = \ \sqrt{d_{AC}^2 \ + \ d_{BC}^2 \ - \ 2 \times d_{AC} \times d_{BC} \times \cos \theta} \ = \ \sqrt{5^2 \ + \ 22.151^2 \ - \ 2 \times 5 \times 22.151 \times \cos 40^\circ} \ = \ \boxed{18.6 \ m} {/eq}

Assuming that the source B is along the horizontal direction (i.e. angle between the two sources is {eq}\theta \ = \ 40^\circ \ deg{/eq}), we find that **speaker A needs to be 18.6 meters away from speaker B.**

#### Learn more about this topic:

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5