# Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase....

## 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?

## Destructive Interference:

When the waves are out of phase during superposition at a point, then the interference is known to be destructive. The destructive interference takes place when the path difference between the wave must be equal to the half of wavelength.

## Answer and Explanation:

**Given Data:**

- Frequency of the wave {eq}\rm (f) = 172 \ Hz {/eq}

- Speed of the sound {eq}\rm (V) = 344 \ m/s {/eq}

Now, the wavelength of the wave would be

{eq}\rm \lambda = \dfrac{V}{f} \\ \lambda = \dfrac{344}{172} \\ \lambda = 2 \ m {/eq}

Since, the distance from the Speaker A is 8 m which is the integral multiple of the wavelength, therefore, we can say if at this point, the distanc of the speaker B must be half of the wavelength.

hence the closest distance of the speaker B would be

{eq}\rm d = \dfrac{\lambda}{2} \\ d = \dfrac{2}{2} \\ d = 1\ m {/eq}

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from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5