# Simple Harmonic Motion, Initial Displacement vs Initial Cond

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1. Feb 20, 2015

### austrosam

Hi everybody,

I'm writing an exploration on the mathematics of simple harmonic motion and I stumbled across something I fail to understand in one of my resources (http://tutorial.math.lamar.edu/Classes/DE/Vibrations.aspx). In the example the author uses toward the end of the resource, the object is initially displaced by 6 inches (don't ask me why he felt the need to use imperial units) but then, the initial condition for displacement given at t=0 is -1/2. Should it not be 6?

My guess is that one can simply set t=0 at any point during the oscillation and not in fact when the oscillation is started, but that still would not quite explain everything. Maybe I am just being very silly...

Sam

2. Feb 20, 2015

### CWatters

If you pull a pendulum to one side by 6" and let it go it will swing back and forth a total distance of 12". So the initial displacement is half the peak to peak amplitude. It sounds like they decided to use "fraction of peak to peak amplitude" as the unit of displacement rather than inches or meters.

The minus sign is probably because the restoring force is in the opposite direction to the displacement.

3. Feb 20, 2015

### austrosam

Right, I guess that makes sense, though I must say it still seems a little odd, to me it would seem much more straightforward to use a value of 6 inches instead.

Many thanks anyway!

One more thing, I merely need quick confirmation I'm on the right track here. Later on, they calculated the amplitude which was slightly larger than the initial displacement. Is this because of the initial velocity, and the object not starting from rest?

4. Feb 20, 2015

### bhillyard

I f the author is using imperial units then the lengths would be in feet - so 6 inches is 1/2 a foot. Then they would use g=32 ft/sec/sec.

5. Feb 20, 2015

### austrosam

Perfect!!! Thank you! I am totally unfamiliar with imperial units, I should have really checked that. Thanks!!