A particle of mass ,m, is at rest at the end of a spring(force constant=k)hanging from a fixed support. At t=0 a constant downward forcd F is applied to the mass and acts for a time T. Show that after the force is removed, the displacement of the mass from its equilibrium position (x=Xe, where x is down) is:

x - Xe =F/k[cosW(t-T) - cosw(t)]

where w = omega

Solution Summary

This solution shows that after the force is removed, the displacement of mass from equilibrium position is x - Xe =F/k[cosW(t-T) - cosw(t)]. Steps are shown for the full derivations for where the concepts of Hooke's Law and period are used.

(Please see the attached file).
Two small blocks, A and B , of masses 0.8 kg and 1.2 kg respectively, are stuck together. A spring has natural length 0.5 meters and stiffness of 98 N/m. One end of the spring is attached to the top of the block A and the other end of the spring is attached to a fixed point O.
(a) The system

Please see attachment. Thanks.
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Many real springs are more easily stretched than compressed. We can represent this by using different spring constants for and for . As an example, consider a spring that exerts the following restoring force:
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Excel Spreadsheet Problem
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