- #1
bullet_ballet
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"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?
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?