Compensating for Doppler Effect in Moving Car: Formula and Graphing

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SUMMARY

The discussion focuses on calculating the frequency compensation for the Doppler effect experienced by a stationary observer as a car moves past at 50 mph. The relevant formula derived is ƒ_{source} = (v + V_{s})/v * ƒ_{observed}, where V_{s} is the speed of the car. The user initially struggled with the angle θ and its relationship to distance d and time t, but successfully adjusted the expression for V_{o} to V_{o} = (V_{s}^{2}t)/√(d² + V_{s}^{2}t²), leading to a correct graphing of the observed frequency. The final approach confirmed that the graph of V_{o} versus time aligns with expectations.

PREREQUISITES
  • Understanding of the Doppler effect and its mathematical representation
  • Familiarity with trigonometric functions, particularly sine and cosine
  • Basic knowledge of graphing functions and interpreting results
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Explore advanced applications of the Doppler effect in various fields, such as astrophysics
  • Learn about graphing techniques for trigonometric functions in physics
  • Investigate the impact of varying speeds on the Doppler effect using simulation software
  • Study the relationship between frequency, wavelength, and speed in wave mechanics
USEFUL FOR

Students in physics, engineers working with sound and motion, and anyone interested in understanding the practical implications of the Doppler effect in real-world scenarios.

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Homework Statement


Trying to find the formula to generate a sin wave that would compensate for the Doppler effect if played from a car moving 50 mph past a stationary observer 1 meter from the car's path.

Homework Equations


ƒ_{observed} = \frac{v}{v+v_{s}}ƒ_{source}

The Attempt at a Solution


Tried to work this out using variables first. Say d is the distance from observer to car's path.

First, we want to keep the observed frequency constant, so rewrite Doppler formula for source:
ƒ_{source} = \frac{v+v_{s}}{v}ƒ_{observed}

Then, taking the component of the car's velocity towards the observer
V_{o} = V_{s}cosθ

Where θ is the angle between the car's path, and the direct line of sight to the observer.

But we want this in terms of d, time t and Vs, so we can rewrite θ thusly

θ=tan^{-1}(\frac{d}{V_{s}t})

And then plugging back into Vo, we get
V_{o}=\frac{V_{s}}{\sqrt{(\frac{d}{V_{s}t})^{2}+1}}

So plug this back into our Doppler equation.

ƒ_{s}=\frac{v+\frac{V_{s}}{\sqrt{(\frac{d}{V_{s}t})^{2}+1}}}{v}ƒ_{o}

I've tried graphing this using ƒobserved=440 Hz and Vs=22 m/s, and the graph is symmetrical about t = 0, when it obviously should not be. I'm not sure where I'm going wrong.
 
Last edited:
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Check your expression for ##\theta## in terms of ##d##, ##V_s##, and ##t##. Are you sure that's what you want?
 
Why wouldn't it be? Tan gives me opposite and adjacent components which are d and tVs. Are you saying it should be sin or cos?
 
Consider the sign of your expression for ##V_o## as you pass ##t=0##.
 
Am I approaching this the right way? I don't see what else theta can be written as.

would writing it in terms of cosine make sense? then V_{o}=\frac{V_{s}^{2}t}{\sqrt{d^{2}+V_{s}^{2}t^{2}}}
 
Last edited:
Your approach is fine, but you have to be a bit careful. Try plotting ##V_o## vs ##t##. Is it what you expect?
 
Yes! this new Vo works. Thanks!
 

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