Question

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from the speaker.
Take the speed of sound in air to be 344 m/s. What is the closest you can be to speaker B and be at a point of destructive interference?

Answer 1

Answer:

The distance of the speaker B would be 1 meters.

Explanation:

Given that,

Frequency of the waves emitted, f = 172 Hz

You are 8.00 m from the speaker.

The speed of sound in air to be 344 m/s.

The wavelength of the wave is given in terms of frequency and the speed of sound. It is given by :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344\ m/s}{172\ Hz}\\\\\lambda=2\ m$

Since, the distance from the Speaker A is 8 m which is the integral multiple of the wavelength. Let the closest distance of the speaker B would be given by :

$d=\dfrac{\lambda}{2}\\\\d=\dfrac{2}{2}\\\\d=1\ m$

So, the distance of the speaker B would be 1 meters. Hence, this is the required solution.