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Simple Harmonic Motion

  • Thread starter Ed Aboud
  • Start date
  • #1
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Homework Statement


A particle moves with simple harmonic motion of period [tex] \frac{\pi}{2} [/tex]. Initially it is 8cm from the centre of motion and moving away from the centre with a speed of [tex] 4 \sqrt{2} [/tex] cm/s.
Find an equation for the position of the particle in time t second.


Homework Equations


[tex] x = A \cos{ \omega t + \epsilon} [/tex]
[tex] v^2 = \omega^2 (A^2 - x^2) [/tex]
[tex] T = \frac{2 \pi}{\omega} [/tex]

The Attempt at a Solution


[tex] T = \frac{2 \pi}{\omega} [/tex]
[tex] \omega = 4 rad s^-1 [/tex]
[tex] v^2 = \omega^2 (A^2 - x^2) [/tex]
[tex] 32 = 16(A^2 - 64) [/tex]
[tex] A = \sqrt{66} [/tex]
[tex] x = A \cos( \omega t + \epsilon) [/tex]
[tex] x = \sqrt{66}\cos(4t + \epsilon) [/tex]

The answer in the book is:
[tex] x = \sqrt{66}\cos(4t + .175) [/tex]

I don't understand where the .175 comes from.
Thanks for any help.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
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Make use of the initial conditions at t = 0 to solve for the phase.
 
  • #3
199
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Thanks for helping.
So at t=0 x=8

[tex] 8 = \sqrt{66}\cos(\epsilon) [/tex]
[tex] \cos(\epsilon) = \frac{8}{\sqrt{66}} [/tex]
[tex] \epsilon = \arccos\frac{8}{\sqrt{66}} [/tex]
[tex] \epsilon = 10.02498786 [/tex]

Have I made a mistake somewhere?
 
Last edited:
  • #4
Doc Al
Mentor
44,882
1,129
Use radians, not degrees. :wink:
 
  • #5
199
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God, that is embarrassing ha.
Thanks very much.
 

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