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Simple harmonic motion,

  1. Feb 25, 2006 #1
    I am trying to solve the part c of the question. I know the initial condidtions of the motion which are Xo=3 and Vo=50 my W=10 , B1=3 B2=5 and my equation is
    x(t)=3cos(10t)+5sin(10t), the question is at which first times x=0 and x(2dot)=V=0,
    Now, when I plot the equation on excel, I clearly see that at Xmax (Amax) V=0 at t=0.1 and when V=Vmax x=0 at around t=0.26s.

    But when I want to find these times by solving the equation , I find something different.
    how to solve this equation to find the first times at wich x=0 and v=0, what am I missing while solving it ?

    anyhelp appreciated

    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2
    What do you mean by:
    ? Are you actually subtracting the two?

    Solving trig equations can be a real pain. There are a few guidelines, though, that can make it a bit easier.
    1) Try to get all of the trig functions to the same argument. That's already done here.
    2) Try to get all the trig functions to a single trig function. (This is the hard one...more below.)
    3) Recall the limits of the trig functions, as they might help you decide where to focus your attention for zeros.
    4) Have enough sanity to realize when and if it won't work and do it numerically! :)

    Not much more to advise. I don't know of any hard and fast rules for solving them.

    4) is not needed in this case!

    Now, about 2). What I'm talking about here are trig identities. A few of them come up more often than not:
    [tex]sin^2 \theta + cos^2 \theta = 1[/tex]
    [tex]cos(2 \theta)=cos^2 \theta - sin^2 \theta = 2cos^2 \theta -1=1-2sin^2 \theta [/tex]
    [tex]sin(2 \theta)=2sin \theta cos\theta [/tex]
    The half-angle formulas are derived from these, but I haven't had many occasions to use them.

    In your case the first relation looks good. Solve it for the cosine and plug it into your equation. That will give you a second order equation for cosine. Ugly, but it'll work.

  4. Feb 25, 2006 #3
    sorry !! I was spossed to write (+) instead of (=)

    I tried to get squared of (X=....) but since I have coefficients of cos and sine so I cant use the (sin^2 \theta + cos^2 \theta = 1)

    what do you mean with the first one can you explain or give an example ( Sorry English is my second language and sometimes I am having trouble with understanding)

    Last edited: Feb 25, 2006
  5. Feb 26, 2006 #4
    Try solving [tex]sin^2 \theta + cos^2 \theta = 1[/tex] for either sine or cosine. Then plug it into your equation.

  6. Feb 26, 2006 #5
    thanks ,Dan I will try it now :)

    What I wonder is that , this is asking me the times at which X=0 and V=0 for the first time , what if It asked me the times at which X=0 and V=0 for the second time ? How do I solve that with my equation ? We know that the motion occurs every 2Pi/w time but how to use that to solve in this particular stuation.? Do you have any idea?
    thanks again
  7. Feb 26, 2006 #6
    When you work out your solution it will be in terms of a sine or cosine function, depending on which you pick to work with. These functions are periodic so you will get a general solution sort of like [tex]t= \frac{2 \pi n}{3}[/tex] (I made that up) where n is some integer. By letting n be different integers, we get different points in the oscillating cycle.

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