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More Simple Harmonic Oscillations

  1. Nov 8, 2006 #1
    An object in simple harmonic motion oscillates with a period of 4.00 s and an amplitude of 9.08 cm. How long does the object take to move from x=0.00 cm to x=5.07 cm?

    I set up my eqn like this: 0.0908cos(ωt)=0.0507
    then with ω=90deg I get 0.623s which is slightly higher than I think the answer would be, and is ultimately incorrect.

    I've tried working the problem out a number of times, in rad and degs to make sure I'm not making an error there, and I get the same answer.

    The velocity of an object in simple harmonic motion is given by v(t)= -(0.250 m/s)sin(17.0t + 1.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.120 m/s?

    for this one it seems like I just have to work it through to find t. When I do so, I get down to 17.0t+1.00π=28.7 (from using sin^-1=-.120/-.250) and then there is no answer that fits (0<t<5 approx.). Plus, in trying the answers, they are wrong.

    also, I have another problem I posted about here

    I'm pretty spent on these, any and all help would be appreciated.
  2. jcsd
  3. Nov 8, 2006 #2
    You have two condtions x(0)=0, and x(t)=5.07cm. To use the equation that [tex]x(t)=Acos(\omega t + \phi)[/tex] you have to first find the phase constant. So you need to make a statement like
    [tex]x(t)=Acos(\omega t + \phi)=0[/tex]

    It is pretty obvious for this one because you essentially want to just start with a sine function. But you can solve and find the phase. Then you want to use that phase you found to solve when [tex]x(t)=Acos(\omega t + \phi)=5.07cm[/tex]. Another problem I see with what you did is you are saying the angular frequency has units of degrees? Remember that [tex]\omega = \frac{2 \pi}{T} [/tex] where T is the period, and has units of s^-1.

    The second one is very similar, if you can get the first one then you can get the second one.
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