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Simple harmonic motion (with calculus)

  • #1
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Homework Statement


An object with mass m undergoes simple harmonic motion, following 2 perpendicular directions, described by the equations:

x=a cos (wt), a>0,
y=b cos (2wt), b>0

a) find the equation of the trajectory
b) find the speed at any given time (so having t as a variable)
c) the maximum force F which acts on the object at any given time (again, having t as a variable).

Homework Equations




The Attempt at a Solution


So far: a) from x=a cos (wt) we get cos (wt)=x/a; in the y equation, we can expand as follows:
cos (2wt)=cos^2 (wt) - sin^2 (wt). We also know that for any real x we have cos^2 (x) +sin^2(x)=1, therefore cos (2wt)=cos^2 (wt) - sin^2 (wt)=cos^2 (wt) + cos^2(wt)-1; therefore y=b (2cos^2 (wt) -1 )=b( x^2/a^2 -1 ), which is the equation of the trajectory.

Now for b) and c), I'm not quite sure how to use what I have. I differentiated the x and y equation from the beginning, differentiated the trajectory and somehow I need to combine them. I suppose the idea from b) applies to c).

I would be grateful if you could give me some hints :)

Have a great day,
Adrian
PS: If it is not clear, I could rewrite using latex.
 

Answers and Replies

  • #2
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Solved b), only need further help at c) (got that Fmax=mw^2 (8b-4ab)/a , which doesn't seem correct to me )
 
  • #3
haruspex
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Solved b), only need further help at c) (got that Fmax=mw^2 (8b-4ab)/a , which doesn't seem correct to me )
8b-4ab is dimensionally inconsistent, so cannot be right. Step through your working to find where the dimensional inconsistency arises.
 
  • #4
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Apparently, I forgot that since those 2 directions are perpendicular, a= sqrt ( ax^2+ay^2) , where ax is the acceleration found by differentiating vx from b) and the same for ay. Basically, I have to maximize sqrt ( a^2*w^4*cos^2 (wt) + 16*b^2*w^4*cos^2( 2wt) ), which doesn't require calculus, the maximum is obviously achieved when both cosines are =1, and therefore the maximum acceleration would be sqrt (a^2*w^4+16*b^2*w^4). I might be missing sth though, does that seem correct?
 
  • #5
haruspex
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Apparently, I forgot that since those 2 directions are perpendicular, a= sqrt ( ax^2+ay^2) , where ax is the acceleration found by differentiating vx from b) and the same for ay. Basically, I have to maximize sqrt ( a^2*w^4*cos^2 (wt) + 16*b^2*w^4*cos^2( 2wt) ), which doesn't require calculus, the maximum is obviously achieved when both cosines are =1, and therefore the maximum acceleration would be sqrt (a^2*w^4+16*b^2*w^4). I might be missing sth though, does that seem correct?
There is one step missing in the reasoning above, though perhaps you merely omitted to mention it: that both cosines can be 1 simultaneously.
 
  • #6
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Not sure if I understand your point, by saying "the maximum is obviously achieved when both cosines are =1" I meant to say that the maximum occurs when both cosines are equal to 1 (when t=2k pi, k integer). Is this correct, sir?
 
  • #7
haruspex
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when t=2k pi, k integer
Yes, that's the part you needed to add for completeness. E.g. if the expression had been ##\cos(\omega t)+\cos(\omega t+\cos(t))## it might not be possible for that ever to equal 2.
 
  • #8
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I see your point, thank you for taking your time to help me, haruspex. Wish you an amazing day!
 

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