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Homework Help: Simple Harmonic Motion (SHM)

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the simple harmonic motion of a block attached to a spring on a horizontal surface (assume no damping to begin with). At t = 0, the potential energy in the spring is 25% of the maximum potential energy during this motion. Moreover, the kinetic energy of the block is decreasing with time at t = 0 and, at t = 2.0 s, the kinetic energy becomes zero for the 1st time (after t = 0).

    (i) What is the frequency of this motion?
    (ii) What is the spring constant if the mass of the block is 2 kg?
    (iii) If the total energy of the system is 10 −2 J, then what is the amplitude of the motion?
    (iv) Next, a drag force is added with a damping constant of 0.2 kg/s. At what time will the
    energy in the motion reduce to 25% of the energy at t = 0?


    2. Relevant equations

    F = 1/ T
    x(t) = Acos(2*pi*t/T)
    omega = 2 * pie * frequency
    Xmax(t) = Ae^(-bt/2m) **FOR PART 4

    3. The attempt at a solution

    I drew up the image of the simple harmonic motion and i figured that if i find the time it takes to make one oscillation, i could then find the frequency (F =1/T). However, It feels like a guessing game for the time. According to the drawing, it takes about 3 seconds to make a full oscillation; does that sound right? That would mean the frequency would be .33 hertz.

    For part ii, the spring constant is : k = omega^2*mass. Omega is our unknown and I would assume as 2 * pie * frequency. But im nervous my frequency is wrong.

    For part iii, you can use the equation x(t) = Acos(omega + phase constant). Can I assume the phase constant to be zero? If so, we can use the above formula to solve for A.

    Part 4, we use the above formula and were given the damping constant but I'm confused on the topic of damping oscillations and not sure where to go with the formula.

    All advise, hints, tips, and help will be greatly appreciated. If I can rep people, I will do that as well for help.

    -UMDstudent
     
  2. jcsd
  3. Feb 3, 2010 #2

    vela

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    The potential energy in the spring depends on x(t), so what this sentence is telling you in an indirect way is the possible positions of the block at t=0.

    What's the formula for the potential energy of a spring? The spring stores the maximum potential energy when it is stretched or compressed as much as possible by the block, i.e. when x(t)=A, where A is the amplitude of oscillation, so what is the maximum potential energy equal to in terms of A? Finally, what equation does the sentence translate into?

    If you solve for the initial position, it'll turn out the block could be in any of four spots in the cycle of oscillation. What this sentence is telling you is that the block is slowing down at t=0. This will narrow down the possibilities to just two. That'll give you enough information to figure out what the frequency of oscillation is.

    A period of 3 seconds is way off. What you know is the block took two seconds to go through less than 1/4 of a cycle, so the period has to be greater than 12 s.
     
  4. Feb 3, 2010 #3
    The potential energy of a spring is U = 1/2kx^2. In this case, this formula translates into 1/2kA^2. Theres a problem with this though and that we don't know the spring constant nor do we know the amplitude. How does this help me?
     
  5. Feb 4, 2010 #4

    vela

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    You don't need to know either to figure out where in the cycle it is at t=0 because they'll cancel out.

    One thing I just noticed is that your equation for x(t) is incomplete. You want to use the more general form [itex]x(t)=A\cos(2\pi t/T+\phi)[/itex], and solve for [itex]\phi[/itex]. The angle [itex]\phi[/itex] tells you where in the cycle it was at t=0. At t=2 s, the block comes to rest momentarily, and you should be able to deduce what [itex]2\pi t/T+\phi[/itex] has to equal at that instant, allowing you to solve for T.
     
  6. Feb 4, 2010 #5
    The 25% throws me off
     
  7. Feb 4, 2010 #6
    What is the point of total energy in iii) ?
     
  8. Feb 4, 2010 #7

    vela

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    You need it to calculate the amplitude.
     
  9. Feb 4, 2010 #8
    So the phase constant is equal to omega (2*pie*frequency) + Phase not. I feel like we keep bring unknowns to this equation and after reading your posts, i'm getting more confused. So what your saying is that the phase constant isn't zero which tells us its some distance away from X at T=0? How do I solve for phase constant?
     
  10. Feb 4, 2010 #9

    vela

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    It sounds to me like you're reluctant to write any equations down because they contain variables you don't know the value of yet, but these problems aren't the type where you write one formula down, fill in a bunch of numbers, and solve for the one unknown variable. They require a bit more work. Sometimes you will immediately see how to solve the problem; other times you won't. When you can't, you start by examining various relationships and see where they lead.

    So how do you solve for the phase constant? You start by doing what I said in my initial post and translating the first sentence into an equation. It said the potential energy at t=0 was 25% of its maximum. You told me the potential energy is (1/2)kx(t)^2 and the maximum is (1/2)kA^2, so you get

    [tex]\frac{1}{2}kx(0)^2 = 0.25\left(\frac{1}{2}kA^2\right)[/tex].

    You also have that [itex]x(0)=A\cos(\phi)[/itex], so you plug that in and you get

    [tex]\frac{1}{2}kA^2\cos^2(\phi) = 0.25\left(\frac{1}{2}kA^2\right)[/tex].

    You don't know k or A, but it doesn't matter because they cancel out. You're left with only one unknown, [itex]\phi[/itex], so you can solve for it. You should get 4 solutions.
     
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