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Simple harmonic motion problem

  1. Dec 6, 2006 #1
    Jane wants to swing on a rope across a river. What minimum speed does she need to make it across, and once she's across, what minimum speed does she need to make it back?

    Here's what's given:
    mass = 47 kg
    horizontal wind - call it F - (opposite to her swing) = 120 N
    horizontal distance (D) = 50 m
    rope length (L) = 40 m
    theta = 50 degrees

    (hopefully you can see the image)

    Here's how I started:

    D = Lsin(theta) + Lsin(phi)

    Plug in the values, and phi = 28.9 degrees

    Then, Change in height = Lcos(phi) - Lcos(theta)
    Plug in the values, change in height = 9.3 m

    From there:

    PE(o) + KE(o) + wind = PE(f) + KE(f)

    mgh(0) + (1/2)mv(0)^2 - F(w)*D = mgh(f) + (1/2)mv(f)^2

    and v(0) = 8.55 m/s

    I got that part right. But, now how do I find the minimum velocity to go back? I tried switching the h(0) and h(f) in that last equation to go back, but it didn't work. What's wrong?
  2. jcsd
  3. Dec 7, 2006 #2


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    Have you considered that she is now swinging with the wind?
  4. Dec 7, 2006 #3
    Yes, I did. Here's what happens:

    mgh(0) + (1/2)mv(0)^2 + F(w)*D = mgh(f) + (1/2)mv(f)^2

    46(9.8)(-9.3) + (1/2)(47)v(o)^2 + 120(50) = 0

    and v(o)^2 = -76.92

    But the fact that it's negative makes me think that it's wrong. Can I still take the square root of it?

    Edit: Apparently I can. I just tried it again and got it right.
    Last edited: Dec 7, 2006
  5. Dec 7, 2006 #4
    But if you got a negative velocity, it means that the wind provides enough energy to Jane to reach that point, so that she doesn't need any initial velocity.
  6. Dec 7, 2006 #5
    That's weird. Are you sure the negative doesn't just mean that Jane is swinging in the opposite direction? Is there something I should have done to make the velocity squared positive?
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