Determine the particle's equation of motion

  • Thread starter Sheneron
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  • #1
Sheneron
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Homework Statement


A particle with a mass of 0.500 kg is attatched to a spring with a force constant of 50.0 N/m. At time t = 0 the particle has its maximum speed of 20.0 m/s and is moving to the left. (Use t as necessary.)
(a) Determine the particle's equation of motion, specifying its position as a function of t, time.
(b) Where in the motion is the potential energy three times the kinetic energy?
(c) Find the length of a simple pendulum with the same period.
(d) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m.

The Attempt at a Solution


I only need help with the first part for now, but here is what I tried.

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
[tex]\omega = 10[/tex]

[tex]v_{max} = A\omega[/tex]
[tex]A = 2[/tex]
So my equation of motion looked like this.
[tex]x(t) = 2cos(10t)[/tex]

but that was wrong and I am not entirely sure why
 

Answers and Replies

  • #2
Doc Al
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You are off by a phase factor. Note that at t = 0 the particle is at max speed--so what's its position with respect to equilibrium?
 
  • #3
Sheneron
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It's position, with respect to equilibrium, would be 0, which means the phase factor needs to be pi?
 
  • #4
Doc Al
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It's position, with respect to equilibrium, would be 0,
Right.
which means the phase factor needs to be pi?
Write the general expression and see if that works.
 
  • #5
Sheneron
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Ok that worked and I am on part D) now.

x(t)= 2sin(10t + pi)

I set x to 1 and solved for t but that didn't work. Can you not do that?
 
  • #6
Doc Al
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That should work fine. What did you get? (Remember that you're working in radians.)
 
  • #7
Sheneron
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[tex]\frac{sin^{-1}(.5)-\pi}{10}[/tex]


that gives me a negative .261

perhaps I am not doing my algebra right?
 
  • #8
Doc Al
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Two things you need to do:
(1) Choose a value of [itex]sin^{-1}(.5)[/itex] that gives you a positive time.
(2) Figure out the times when x = 0. (t = 0 is one such time, but it's not the one you want. At t = 0 the particle is moving left; you want the time when it's moving right.)
 
  • #9
Sheneron
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but my time = all that stuff, and sin of .5 - pi always comes out a negative number..
 
  • #10
Doc Al
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but my time = all that stuff, and sin of .5 - pi always comes out a negative number..
Realize that there's more than one angle that satisfies [itex]\sin\theta = .5[/itex].
 
  • #11
Sheneron
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For arcsin of 0.5 I get a value of 0.52359, so you are saying there is another value in addition to this?
 
  • #12
Sheneron
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Can anyone that is still awake help me solve part D?
 
  • #13
Doc Al
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For arcsin of 0.5 I get a value of 0.52359, so you are saying there is another value in addition to this?
I think you've figured it out by now, but sure:
[tex]\sin\theta = \sin(\theta + 2\pi)[/tex]

That's just one example (but it's the one you need). Look at a graph of a sine function and see all the places where it equals .5.
 
  • #14
Sheneron
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I figured it out, but I still didn't understand it, so I am glad you posted again.
 

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