# Sound with a frequency of 1210 Hz leaves a room through a doorway with a width of 1.00 m. At...

## Question:

Sound with a frequency of {eq}1210\ Hz {/eq} leaves a room through a doorway with a width of {eq}1.00\ m {/eq}. At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use {eq}344\ \dfrac ms {/eq} for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.

## Diffraction Effects in Sound

Sound waves also gets diffracted when it passes through openings. The sound waves bend at the openings and the bend waves interfere with direct waves of sound and produce diffraction pattern of maxima and minimum intensity of sound. The condition for getting diffraction maxima and minima of sound is just same as that in the case of light. If the path difference is an integer multiple of wavelength we get constructive interference or larger sound and if the path difference is an odd multiple of half wavelength destructive interference or minimum sound.

## Answer and Explanation:

**Given points**

- Frequency of the sound wave
*F*= 1210 Hz - Width of the door through which the sound comes out
*d*= 1.0 m - Speed of sound wave
*v*= 344 m/s

Wavelength of the sound wave {eq}\lambda = \dfrac { v } { F } \\ \lambda = \dfrac { 344 } { 1210 } \\ \lambda = 0.284 \ \ m {/eq}

The sound coming out of the door will be zero at points where destructive interference happens.

The condition for destructive interference is {eq}d \sin \theta_n = ( 2 n + 1 ) \times \dfrac { \lambda } { 2 } {/eq}

Here {eq}n, \ \ \theta_n {/eq} are the order of the minima and angle of deviation from the central maximum to the *n* th minimum.

so we can get the angles on either side of the central line several angles for which diffraction minima of sound happens by giving values to *n* as 0, 1, 2, 3, ... etc.

The first minimum happens at some angle when *n* =0

Then we get {eq}d \sin \theta_0 = \dfrac { \lambda } { 2 } {/eq}

So the angle corresponding to first minimum {eq}\theta_0 = \sin ^{-1 } ( \dfrac { \lambda } { 2 d } ) \\ \theta_0 = \sin^{-1 } ( \dfrac { 0.284 } { 2 \times 1 } ) \\ \theta_0 = 8.16^o {/eq}

When *n* = 1 we get the next angle on either side of the central line we get the next minimum sound.

So the angle from the central line for the next minimum {eq}\theta_1 = \sin ^{-1 } ( \dfrac { 3 \lambda } { 2 d } ) \\ \theta_1 = \sin^{-1 } ( \dfrac { 3 \times 0.284 } { 2 \times 1 } ) \\ \theta_1 = 25.21^o {/eq}

The next angle of minimum intensity can be obtained by giving *n* = 2

Then the angle of next minimum intensity {eq}\theta_2 = \sin ^{-1 } ( \dfrac { 5 \lambda } { 2 d } ) \\ \theta_2 = \sin^{-1 } ( \dfrac { 5 \times 0.284 } { 2 \times 1 } ) \\ \theta_2 = 45.23 ^o {/eq}

The next angle of minimum intensity can be obtained by giving *n* = 3

Then the corresponding angle of minimum intensity {eq}\theta_3 = \sin ^{-1 } ( \dfrac { 7 \lambda } { 2 d } ) \\ \theta_3 = \sin^{-1 } ( \dfrac { 7 \times 0.284 } { 2 \times 1 } ) \\ \theta_3 = 83.72^o {/eq}

These are the angles on either side of the central line where we will get minimum intensity of sound.

No further order is possible. Because the height order diffraction possible is {eq}n_m= \dfrac { d } { \lambda } \\ n_m = \dfrac { 1 } { 0.284 } \\ n_m = 3.52 {/eq}

#### Learn more about this topic:

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5