Compensating for Doppler Effect in Moving Car: Formula and Graphing

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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


[itex]ƒ_{observed} = \frac{v}{v+v_{s}}ƒ_{source}[/itex]

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:
[itex]ƒ_{source} = \frac{v+v_{s}}{v}ƒ_{observed}[/itex]

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

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

[itex]θ=tan^{-1}(\frac{d}{V_{s}t})[/itex]

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

So plug this back into our Doppler equation.

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

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.
 
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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?
 
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 [itex]V_{o}=\frac{V_{s}^{2}t}{\sqrt{d^{2}+V_{s}^{2}t^{2}}}[/itex]
 
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Yes! this new Vo works. Thanks!