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Finding velocity.

  1. Nov 10, 2004 #1
    Ok this problem does have me stumped. This is the area in the chapter I am a little sketchy on. So here is the prob.

    A .5kg mass at the end of a spring vibrates at 3hz, with an amplitude of .15m. What is the velocity when it passes through the equalibrium point.

    So I figured that this is an energy situation where E=1/2mv^2. But do I know the energy? Also where does the frequency come in?
    Thanks again
    Chris
     
  2. jcsd
  3. Nov 10, 2004 #2

    jamesrc

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    If you want to consider the energy think of it this way:

    In this (lossless) system, potential energy is being converted into kinetic energy and back again in such a way that the total energy is constant. You can calculate the kinetic energy like you said: K = .5mv2

    The potential energy is given by U = .5kx2, where x is the displacement from the equilibrium position.

    You should have seen somewhere that the natural frequency of the system is given by [tex]\omega^2 = \frac k m [/tex], so that you can find k from given quantities.

    Consider 2 locations of the mass:

    1st at the equilibrium position, the potential energy is zero. This makes intuitive sense since the spring is not compressed or extended and follows from the equation we just saw. At this position, your equation holds true: E = .5mv2, where v is what we're trying to find.

    The 2nd position we should consider is where the mass is at the edge of its travel: x = A = .15 m. At this point, what is the kinetic energy of the mass (hint: the mass stops at this point as it changes direction). Now you can calculate E again.

    Set your two expressions for E equal to each other and solve for v.

    (The other way to do it is if you know the equation for SHM position as a function of time --> velocity as a function of time.)
     
  4. Nov 10, 2004 #3
    you could set this up something like this:

    x(t) = A sin(wt)

    where w = 2*pi*f and A is the Amplitude.

    velocity is given by dx/dt

    Now, you know that at the equilibrium point, the velocity will be a maximum, so there's a few ways to go about doing it from here.

    I would find what values of t give x=0. (remember that sin(m*Pi) = 0 , m=0,1,2...etc)

    And then substitute into the velocity equation.
     
    Last edited: Nov 10, 2004
  5. Nov 11, 2004 #4
    jamesrc I think I tried what you were saying and I still got the wrong answer.
    I did [tex] 1/2 m v ^2 = 1/2 k x^2 [/tex] to find k I used [tex] f = 1/2 \pi \sqrt k/m [/tex] Then x is eqal to A when all the energy is potential. I just don't see what I did wrong???

    k=177.65, x=.15, and I find v to be equal to 7.30034
    thanks
     
  6. Nov 11, 2004 #5
    V=2*pi*f*A, where f-frequency, so v=3m/s
    Isn't it correct?
     
  7. Nov 11, 2004 #6
    With the way I would do it, I got v=2.8 m/s, which with sig figs v=3 m/s

    same as Yegor
     
  8. Nov 11, 2004 #7

    jamesrc

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    compute_a_nerd: Either way you do it, you should find that [tex] v = A \omega [/tex], so you may just want to check your algebra; (without skipping too many steps):

    [tex] \frac{mv^2}2 = \frac{kx^2}2 [/tex]
    [tex] v = \sqrt{\frac{kx^2}m} [/tex]
    [tex] v = A\sqrt{\frac k m } [/tex]
    [tex] v = A\omega [/tex]

    which is the same numerically as what the other guys here were saying.
     
  9. Nov 11, 2004 #8
    Thanks guys your right, just a plug and chug mistake. I must have been out of it last night, cause I just couldn't get it to work. lol.
    thanks a lot
    Chris
     
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