Calculating Maximum Net Force in Simple Harmonic Motion

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Dark visitor, do you see the similarity between this problem and the engine problem. There we were looking for max velocity. So when you differentiate position with respect to t, the w pops out in front (chain rule) and sin becomes cos. D/dx again and you get -w*w*sin.

Next time you do this, please have all your notes out and for good heavens, my man, don't try to bite off so much in a weekend.:eek: Physics requires slow assimilation and lots of practice!
 
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Only one more thing, sorry.

Is the amplitude (A) .160 m? And is [tex]\omega[/tex] equal to[tex]\pi[/tex]/16?
 
Dark Visitor said:
Only one more thing, sorry.

Is the amplitude (A) .160 m? And is [tex]\omega[/tex] equal to[tex]\pi[/tex]/16?

yes.
 
Okay, thanks. Here is my work:

amax = ([tex]\pi[/tex]/16)2(.160 m)
= .00617 m/s2

Fmax = (.64 kg)(.00617 m/s2)
= .003949

or 3.9 * 10-3

Thanks a lot. I really appreciate all of your help.