Find Resoance Frequency for Electric Motor Mass of 100 kg

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SUMMARY

The discussion focuses on calculating the resonance frequency for an electric motor with a mass of 100 kg supported by vertical springs that compress by 1 mm. The correct formula for resonance frequency is derived from ω = √(g/x), where g is the acceleration due to gravity (9.82 m/s²) and x is the spring compression (0.001 m). The initial miscalculation led to an incorrect frequency of 5950 rpm, but after correcting the spring compression value, the accurate resonance frequency is determined to be approximately 945 rpm. This highlights the importance of precise unit conversion in mechanical calculations.

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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?
 
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you used x = 1 mm as if it was x = 1 m.
 
suffian said:
you used x = 1 mm as if it was x = 1 m.

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.
 
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.
 
Thanks a lot.
 

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