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Q98P

Expert-verifiedFound in: Page 512

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?**

- Adjacent wave front separation is$0.5\text{m}$
- Wave front separation in front of source is $0.34\text{m}$
- Wave front separation behind the source is $0.66\text{m}$

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

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

Formula:

The wavelength shift in the interference pattern$\Delta \lambda ={\lambda}_{0}\frac{(\pm v)}{c}$(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)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 $0.5m$

** **

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$0.34\text{\hspace{0.17em}m}$

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 $0.66\text{\hspace{0.17em}m}$

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