# Two fire alarms are placed in a building, one 2.3 m north of the other. A person sitting at a...

## Question:

Two fire alarms are placed in a building, one 2.3 m north of the other. A person sitting at a desk which is 67.2 m east and 8.4 m north of the point mid-way between the alarms does not hear them when they sound. Speed of sound in air=330m/s.

What is the lowest frequency that the alarms can be sounding at assuming that the alarms are both perfectly in phase?

## Constructive and Destructive interference:

Constructive and destructive interference consists of the interpolation of waves, such as when the crests line up, the amplitude will increase by adding the amplitude of the two waves. This is defined as constructive interference. On the other hand, when opposing crests line up, the amplitude of the 2 waves will cancel out, resulting in destructive interference. There will be a relationship between the distance of the 2 sources and the wavelength of the interference.

• Constructive interference:

{eq}\Delta x=n\lambda {/eq}

• Destructive interference:

{eq}\Delta x=n\dfrac{\lambda}{2} {/eq}

The lowest frequency will be at the first point of interference between the two speakers. First, we need to determine the actual distance between the person and the two speakers. We use trigonometry to find the actual distance of each speaker to the person.

• Distance of the first speaker:

{eq}d_1=\sqrt{d_x^2+d_y^2}\\ d_1=\sqrt{67.2^2+(8.4-1.15)^2}\\ d_1=67.59\ m {/eq}

• Distance of the second speaker:

{eq}d_2=\sqrt{d_x^2+d_y^2}\\ d_2=\sqrt{67.2^2+(8.4+1.15)^2}\\ d_2=67.88\ m {/eq}

We determine the distant difference between the two speakers and the person:

{eq}\Delta d=d_2-d_1= 67.88-67.59=0.29\ m {/eq}

As destructive interference occurs at the point where the person is, we can define that:

{eq}\Delta d=n\dfrac{\lambda}{2}\\ \lambda=2\Delta d\\ \lambda=0.58\ m {/eq}

Finally, we can calculate the frequency by using the speed of sound and the wavelength of the speakers with the following relationship.

{eq}v=\lambda f\\ f=\dfrac{v}{\lambda}\\ f=\dfrac{330}{0.58}\\ f=569\ Hz {/eq}