# Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 572 Hz. A...

## Question:

Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of {eq}572\ Hz {/eq}. A microphone is moved between the speakers to record the wave amplitude at points along that line. What is the spacing between points where constructive interference occurs? The speed of sound in air is {eq}343\ m/s {/eq}.

## Constructive and Destructive Interference:

Two in-phase or, more generally, coherent sources of the same frequency will produce an interference pattern in space, consisting of series of maxima and minima of intensity. The condition of an intensity maximum is that the path length difference between the point of observation and two sources of waves equals the integer number of wavelengths.

## Answer and Explanation:

Let us take the distance between two loudspeakers to be *d*. Then the condition for the constructive interference can be stated as follows:

{eq}(d - x_m) - x_m = m \lambda {/eq}

Here

- {eq}x_m {/eq} is the distance from the left speaker corresponding to the {eq}m^{th} {/eq} order of maximum;

- {eq}\lambda = \dfrac {v}{f} {/eq}

is the wavelength of sound;

- {eq}v = 343 \ m/s {/eq} is the speed of sound;

- {eq}f = 572 \ Hz {/eq} is the frequency of sound;

For the distance to the {eq}m^{th} {/eq} maximum from the left speaker, we obtain:

{eq}x_m = \dfrac {d - m\lambda}{2} {/eq}

The distance between consecutive maxima are then given by:

{eq}\Delta x = x_{m + 1} - x_m = \dfrac {\lambda}{2} = \dfrac {v}{2f} {/eq}

Calculating, we obtain:

{eq}\Delta x = \dfrac {343 \ m/s}{2 \cdot 572 \ Hz} \approx 0.3 \ m {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16