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Energy and oscillations combined

  1. Apr 9, 2014 #1
    Hello, I would like some help with the following question. I highly appreciate your help !

    A mass falls freely for 2.0 m before it hits a vertical spring. The mass sticks to the
    spring, and the system begins to oscillate with a period of 2.2 seconds. If you start a
    stopwatch when the mass is at the lowest point of the oscillation (and you set x = 0
    at that point), where is the mass when the stopwatch reads 1.0 seconds?


    So, what I did is that I found (Phi not ø) to be ∏ ( pi ) but I am not very sure.

    I then used period to get ω. ω=2∏/T. Also, ω=√k/m . Therefore I could find the ratio k/m

    Then I used energy conservation so : mg(h+A) = 1/2 kA 2

    I then rewrote it as k/m = ( 2g(h+A))/ A2

    So I got the the amplitude. Then, I used x=Acos(wt+ø)

    I got the final answer to be 3.55 meters above the lowest point. I don't care about numbers as much as I care to get the concepts right and that I got ø right because I always have some trouble with finding it. Thank you !
     
  2. jcsd
  3. Apr 9, 2014 #2

    BiGyElLoWhAt

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    why is [itex]\Phi_{0} = \pi[/itex]
    think about the implications here. at t=0, with [itex]\Phi_{0} = \pi[/itex] what is x = to? it's -A, which can't be the case. Do you see why?
    I think pi/2 sounds better (for a cos oscillation, you could just as easily use sin), but to make sure, this is SHO, right? If the spring starts at equilibrium, it's position should be 0 at t =0. Make sense? So if you're using cos, you want pi/2, and then define your coordinates accordingly (or vice versa). In this situation, down is negative and up is positve. Other than that, I follow it pretty well. The phase shift shouldn't affect anything except your final anser, so just rething that, and I think you're good.
     
  4. Apr 9, 2014 #3
    I thought lowest point means x = A. But I chose to make down as negative, so would this make x = -A ?
     
  5. Apr 9, 2014 #4

    BiGyElLoWhAt

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    well yea, if you're calling equilibrium 0, and positive is up, then everything above equilibrium needs to be >0 and everything under needs to be <0 , just like any standard coordinate system. I got the -A from plugging pi in for phi_0 and 0 for t. and for the record i'm calling t=0 the instant the block hits the spring.
     
  6. Apr 9, 2014 #5

    BiGyElLoWhAt

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    for the lowest point the magnitude of x =A, maybe that's what you're thinking
     
  7. Apr 9, 2014 #6
    Thank you. now I understand that it can't be ∏ because the question said to set the lowest point as x=0. But I still don't understand how to find phi_0 . I have serious trouble with determining phi_0 LOL
     
  8. Apr 9, 2014 #7

    BiGyElLoWhAt

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    The methodical way involves taking your final function, say Acos(wt+ phi) plugging in t=0, and solving for phi using initial conditions. So if you set the lowest point to be zero, then equilibrium is A, and the highest point is 2A. So at t=0 what's the position? Its A, at the next important time, its 0 then A, then 2A. So what your doing then is giving the function a midline. Think about a cos wave with amplitude A and the lowest point is 0 and highest is 2A. You want Acos(wt + phi) + A. The + A is your midline. Now look at initial conditions again (I think I digressed a bit). At t=0, x=A,so cos(wt + phi) has to equal 0. Well since t is 0 you're really looking at cos(phi) which is equal to 0 at 2 points, π/2 and 3π/2. Well now look what happens next. Does x increase or decrease? It approaches 0, so it decreases. At which of these phi's does cos have a negative derivative? That's π/2. After it hits 0 it goes towards -1. Does that make sense?
     
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