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Expert-verified Found in: Page 512 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A point source that is stationary on an x-axis emits a sinusoidal sound wave at a frequency of ${\mathbf{\text{686 Hz}}}$ and speed${\mathbf{\text{343}}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$. The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along x, what is the adjacent wavefront separation? Next, the source moves along at a speed of${\mathbf{\text{110}}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$. Along x, what are the wavefront separations (b) in front of and (c) behind the source?

1. Adjacent wave front separation is$0.5\text{m}$
2. Wave front separation in front of source is $0.34\text{m}$
3. Wave front separation behind the source is $0.66\text{m}$
See the step by step solution

## Step 1: Given data:

1. Frequency of wave is $f=686\text{Hz}$
2. Speed of wave $c=343\frac{\text{m}}{\text{s}}$

## Step 2: Determine the concept of the Doppler Effect

Using Doppler’s Effect, we can get the wavefront separation.

Formula:

The wavelength shift in the interference pattern$\Delta \lambda ={\lambda }_{0}\frac{\left(±v\right)}{c}$(i)

Here,

${\lambda }_{0}=\text{wavelenght of source},\phantom{\rule{0ex}{0ex}}c=\text{speed of light,}\phantom{\rule{0ex}{0ex}}v=\text{speed of source}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The frequency of a wave oscillation, $f=\frac{v}{\lambda }$ (ii)

## Step 3: a) Calculate the adjacent wave front separation

With given values in equation (ii), we can easily find the wave front separation by using the formula as:

$\begin{array}{c}{\lambda }_{0}=\frac{v}{f}\\ =\frac{343}{686}\\ =0.5\text{m}\end{array}$

Hence, the value of adjacent wave front separation is $0.5m$

## Step 4: b) Calculate wave front separation in front of the source

In front of the source, the shift in wavelength using equation (i) is given as:

$\begin{array}{c}\Delta {\lambda }_{front}=-\frac{\left(110\right)\left(0.5\right)}{343}\\ =-0.16\text{\hspace{0.17em}m}\end{array}$

The negative sign shows that the shift is opposite in direction to the speed of the sound.

The wave front separation is given as:

$0.5-0.16=0.34\text{\hspace{0.17em}m}$

Hence, the wave front separation is$0.34\text{\hspace{0.17em}m}$

## Step 5: c) Calculatewave front separation behind the source

Behind the source, the shift in wavelength using equation (i) is given as:

The shift is in the direction of the sound:

$\begin{array}{c}\Delta {\lambda }_{front}=+\frac{\left(110\right)\left(0.5\right)}{343}\\ =0.16\text{\hspace{0.17em}m}\end{array}$

The wave front separation is given as:

$0.5+0.16=0.66\text{\hspace{0.17em}m}$

Hence, the wave front separation is $0.66\text{\hspace{0.17em}m}$ ### Want to see more solutions like these? 