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Homework Help: SHM question

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A block, attached to a spring, moves in Simple Harmonic Motion on a horizontal frictionless
    table. The displacement from equilibrium is given by x = xm cos (ω t + φ ) where x is positive to
    the right of the equilibrium position. The frequency of the motion is 3.10 Hz and the amplitude
    is 15.0 cm. At t = 0.00 s, the velocity of the block is 2.70 m/s to the left and the position is to the
    left of the equilibrium position.
    Calculate the phase angle, φ.

    Answer: 1.96 Rad

    I have no Idea what I'm doing wrong. Im almost certain its a stupid mistake

    3. The attempt at a solution





    now I just plugged and chugged it all into x = xm cos (ω t + φ )

    but i got the wrong answer...help please
  2. jcsd
  3. Apr 13, 2008 #2
    Check your expression for V (I'm guess that is the velocity). Remember that
    [tex]V={dx \over {dt}}[/tex]
  4. Apr 13, 2008 #3
    I dont think that works...

    V=dx/dt = x(t)=0.15cos[tex]\phi[/tex]



    sin[tex]\phi[/tex]=56.9/.15 ---> does not work

    Am I doing something wrong?
  5. Apr 13, 2008 #4
    Take the derivative before plugging in t=0. And remember that you're differentiating with respect to t and not phi.
    [tex]v={dx \over dt}=-\omega A sin(\omega t + \phi)[/tex]
    This is where you want to put your givens.
  6. Apr 13, 2008 #5
    I think the formulas you are using, v=w xm and a=w^2 xm are for the MAXIMUM velocity and acceleration.
  7. Apr 13, 2008 #6
    correct...is that a problem?
  8. Apr 13, 2008 #7
    damn...i cant seem to get it:grumpy:
  9. Apr 13, 2008 #8

    D H

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    What makes you think the velocity is at a maximum? What is the velocity as a function of time (hint: see post #4)?
  10. Apr 13, 2008 #9
    If you plug 2.7 m/s into the formula for the maximum velocity it is a mistake, because 2.7 m/s is not the maximum velocity, it is just the initial velocity.
    You should be able to put v=-2.7, t=0, A=0.15 and w=19.5 into the formula I gave you, then solve for phi.
    Also, make sure that your calculator is set to radian mode.
    Last edited: Apr 13, 2008
  11. Apr 13, 2008 #10

    D H

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    That formulae will not yield the correct answer using a calculator. He will have to make sure the result is in the correct quadrant.
  12. Apr 13, 2008 #11
    But I'm not getting the desired 1.96 radians.
  13. Apr 13, 2008 #12
    RIght. Good call DH.
  14. Apr 13, 2008 #13

    D H

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    Hmm. This is not good:
    The reason is that you spoon-fed the OP a formula, and he/she will be able to "plug in the numbers", get a result, and not have the foggiest idea why things work that way. The OP is missing something very basic here. In retrospect, it would have been much better to press the OP to come up with post #4 rather than give it out.
  15. Apr 13, 2008 #14
    I tried that but it doesnt give me the right answer...1.18 rad, I need 1.96 rad
  16. Apr 13, 2008 #15
    Sorry DH. You're right. I should be more careful and keep in mind what the purpose of this forum is. I just get so excited when I actually know how to help someone, I get carried away. And in this case I wasn't even right. I'll watch myself in the future. Thanks.
  17. Apr 13, 2008 #16
    That one you'll have to work out unhip_crayon.

    And DH..I was wondering what OP means?
  18. Apr 13, 2008 #17

    D H

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    OP=Original Poster

    unhip_crayon, Can you see how Pacopag derived the velocity equation? You should be able to derive this yourself. Give it a shot. Understanding how velocity varies with time is central to this problem.
  19. Apr 13, 2008 #18
    I understand how pacopag derived the velocity equation, the thing is, I haven't slept in 20 hrs, so I'm exhausted, I'm going to bed now. Thanks for the help guys. I''ll email my prof to double check that answer.
  20. Apr 13, 2008 #19

    D H

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    You should have got -1.18 radians by calculator rather than 1.18.

    The sine function hits all non-zero values twice over any interval of length 2*pi. The inverse sine function has a range of pi, not 2*pi. When you use the inverse sine to calculate an angle you have to make sure that the value is in the correct quadrant. You need some other information to do this, but the problem has that information. You were told the initial velocity (it has a sign; I don't think you used that info) and enough about the position (left of the origin) to resolve the angle.
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