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Simple harmonic motion problem

  1. Apr 7, 2005 #1
    An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period of 2.40 s and a maximum speed of 46.0 cm/s.

    What is the amplitude of the oscillation?

    i tryed solving for t by using the equation x(t) = A cos (wt + q) with w = angular frequency and q = phase constant A = amplitude, i tried to solve for the 3 variables by using the equatoin, the derivitive of the equation, and the 2nd derivitive of the equation ... unsuccesful.

    is there an equation other than that one that allows me to calculate the simple harmonic motion? please tell me how to do this problem...
  2. jcsd
  3. Apr 7, 2005 #2


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    You know some initial conditions. If it is released from rest, then x'(0) = 0

    You also know that it is released from rest after being stretched to its maximum displacement: so x(0) = A.

    plugging in t = 0

    Acos(w(0) + q) = Acos(q) = A

    cos(q) = 1

    q = 0

    You have yet another bit of info: what does the maximum speed tell you (at what point in SHM is the oscillator moving fastest?)
  4. Apr 7, 2005 #3
    i see what you're saying

    the maximum speed occurs when it is at its equlibrium point, x = 0 right?
    using the first derivitive i get velocity equation, which is:
    v(t) = - A w sin (wt) w = 2.61
    -46 = - A*2.61 sin(2.61*t) <---this is at the eqi. point, going to the left

    but i still have 2 variables in this equation, and i cant seem to solve it with
    x(t) = A cos (wt)
    0 = A cos (2.61 t) <---this is at the eqi. point, going to the left

    how do i solve this set of equations:
    -46 = - A*2.61 sin(2.61*t)
    0 = A*cos (2.61*t)
  5. Apr 8, 2005 #4


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    Your equations are not quite right, because you haven't figured out at what time they are true. You are still using the variable t. Why? The equations are not true for all time. They are only true at a specific instant (when the object is in the middle). At what time is that? How would you figure it out?

    Hint1: you haven't made use of the information given about the period of the motion yet (and by that I mean, you haven't made use of it directly).

    Hint2: The bottom equation, even when corrected, will not be required, for it will essentially contain the information: 0=0, at the right time (when the object is at the equilibrium position).
    Last edited: Apr 8, 2005
  6. Apr 8, 2005 #5
    Look carefully at the equation for v. When is v maximized?
  7. Apr 8, 2005 #6


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    Nah! For what purpose? You should already know when v is at its maximum in SHM. That's what my hint1 was alluding to. Don't mess him up! Just follow my hints, adrian.
    Last edited: Apr 8, 2005
  8. Apr 8, 2005 #7
    Can you explain what you mean by that? :( Isn't it the amplitude that he is looking for, given the period of oscillation?
  9. Apr 8, 2005 #8
    waiiiit a second..period 2.4 s thats going forward and back...
    cut that by half, thats going one way trip
    cut that by half again....would it be..could it be..AT THE EQILIBRIUM POINT!?
    2.4/4 = 0.6

    so the time to arrive at the equilibrium point is...0.6!

    now i see, the bottom equation does reduce to 0 = 0 when i plug in the t in

    -46 = - A*2.61 sin(2.61*t) t = 0.6
    -46 = - A*2.61 sin(2.61*0.6)
    A = 17.57!!!

    SWEETTT thanks for all your help!
  10. Apr 8, 2005 #9


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    Ramollari: I edited it for clarity, to explain what I meant. Didn't mean to be rude. And adrian figured it out too, illustrating the point perfectly.

    I know you were saying it was at a maximum when sinwT = 1,

    wT = pi/2

    solve for T and you get the same answer, and equally valid and more mathematical way of looking at it, so sorry I jumped on you.
    Last edited: Apr 8, 2005
  11. Apr 8, 2005 #10
    This is how I'd go to solve it:

    [tex]|v_{max}| = \omega A = \frac{2 \pi}{T}A[/tex]

    I would just solve for A

    [tex]A = \frac{v_{max}T}{2\pi} = 17.57cm/s[/tex].

    I think this is more elegant, though the way you solved it is also good.
  12. Apr 8, 2005 #11


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    lol yes, I keep editing my posts on you, which is bad. Yes, as I stated above, I agree it is more elegant.
  13. Apr 8, 2005 #12
    em....i need help again

    now i have a graph


    this gives me problems because now i need to find the phase constant

    the problem first ask me to find the amplitude first then phase constant
    im not quite sure if it's safe to assume it reaches 60 at 4 s, since it is not marked out.

    i assumed the time at v = 60 to b4 4s, with the -30 at 0 s, i tried to solve for q by mking 2 equations out of V(t), doesnt seem to help

    my question is, x(t) v(t) a(t) all have amplitude and phase constant in them, how do i find amplitude and phase constant while they define each other

    edit: thanks guys, i just realized that at point t = 7 x(7) = A, or you can use x(1) = -A either one works, i used that to find the phase constant and solved for the amplitude
    Last edited: Apr 8, 2005
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