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Resonance problem

  1. Sep 1, 2004 #1
    "An electric motor of mass 100 kg is supported by vertical springs which compress by 1 mm when the motor is installed. If the motor's armature is not properly balanced, for what revolutions/minute would a resonance occur?"

    I set my frame of reference at the end of the spring. Therefore, F = kx - mg = 0. To get resonance ω must equal ω0 which is √(k/m) or √(g/x). I know g to be 9.82 m/s² and x to be 1 mm. Therefore, √(g/x) = 31.3 rps or 1878 rpm. The book lists 955 rpm, so where did I go wrong?
     
  2. jcsd
  3. Sep 1, 2004 #2
    you used x = 1 mm as if it was x = 1 m.
     
  4. Sep 1, 2004 #3
    I see that I made the mistake of dividing g by .01m and not .001m, but that only makes the answer worse at 5950 rpm. My mistake is definitely more fundamental, but I can't see it.
     
  5. Sep 1, 2004 #4
    sqrt(g/x) = sqrt( [ 9.80 m/s² ]/[ .001 m ] ) = 98.99 rad/s = [ 98.99 rad/s ][ 1/2pi rev/rad ][ 60 s/min ] = 945 rpm.
     
  6. Sep 1, 2004 #5
    Thanks a lot.
     
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