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Simple Harmonic Motion

  1. Nov 12, 2003 #1
    I just need someone to tell me if I've done the problem right since the answer is mysteriously missing

    The question is about a rotating wheel causing a piston to execute SHM inside a cylinder (the piston being connected to the outside of the wheel). The wheel rotates with an angular speed of 12rads-1 and it has a radius of 30cm.

    What will be the distance of the piston from an end position after 0.10s?

    ω = θ/t
    θ = 12x0.1 = 1.2rad

    displacement from equilibrium point = amplitudexCosθ = 0.3xCos1.2 =0.1087m

    so distance from end position = 0.3-0.1087 = 0.19m (2sf)
     
  2. jcsd
  3. Nov 13, 2003 #2
    Well I've read through it and it looks OK to me. Couldn't you just have asked your teacher though?
     
  4. Nov 13, 2003 #3
    School's out and I'm on my study leave...
     
  5. Nov 14, 2003 #4
    Lucky you, I'm in school until almost Christmas. Then I have exams (with very little study leave) in January.
     
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