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On a day when the speed of sound is 354 m/s, a 2000 Hz sound wave impinged on two slits 30 cm...

Question:

On a day when the speed of sound is 354 m/s, a 2000 Hz sound wave impinged on two slits 30 cm apart. At what angle is the first max located?

Angle of first maximum

The sound waves after passing the through narrow slits interfere with each other and produce an interference pattern in the screen with alternating maximum and minimum frequencies of sound with a central maximum.

Answer and Explanation:

Given:

  • Distance between two slits = d = 30 cm = 0.3 m
  • Speed of sound = v = 354 m/s
  • Frequency of sound = f = 2000 Hz

We need to find the wavelength of the sound wave:

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

{eq}\rm \lambda = \dfrac{354}{2000} {/eq}

{eq}\rm \lambda = 0.177 \ m {/eq}

Then for the first maximum because of interference, the condition is:

{eq}\rm d sin(\theta) = \lambda {/eq}

{eq}\rm \theta= sin^{-1}(\dfrac{\lambda}{d}) {/eq}

{eq}\rm \theta= sin^{-1}(\dfrac{0.177}{0.3}) = 36.15 ^{\circ} {/eq}


Learn more about this topic:

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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
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