Two loudspeakers are placed 4.5 m apart. They produce the same sounds, in step, across a...

Question:

Two loudspeakers are placed 4.5 m apart. They produce the same sounds, in step, across a frequency range of 744 Hz to Point P is located from one loudspeaker and from the other. What frequency of sound from the two speakers will produce destructive interference at point P?

Interference of Sound:

When two sound wave travel simultaneously in a medium the resultant intensity of sound at some place increases and decreases at the other. The redistribution of energy of waves due to superposition is called interference of waves. The points of maximum intensity are called maxima and it occurs when the path difference is an integral multiple of wavelength. The points of minimum intensity are called minima and occur when the path difference is odd multiple of half of wavelngth.

Answer and Explanation:

Given

Distance between two loudspeaker, {eq}D=4.5\ m {/eq}

Frequency {eq}f=744\ Hz {/eq}

Suppose point P is located at a distance 'a' m apart from one loudspeaker and 'b' m from the other, the path difference

{eq}\Delta x=a-b {/eq}

For destructive interference , {eq}\Delta x=(2n+1) \dfrac{\lambda }{2} \\ \Delta x=(2n+1) \dfrac{v }{2f } \\ \Delta x=(2n+1) \dfrac{344 }{2 \times 744 } \\ {/eq}

Suppose path difference is 1.7 m

{eq}1.7=(2n+1) \dfrac{344 }{2 f} \\ {/eq}

For n=4 we find

{eq}f=911\ Hz {/eq}


Learn more about this topic:

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Interference Patterns of Sound Waves

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5
3.9K

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