The spring mass equation for free motion is
mdt2d2x=−kx Hooke’s Law
mg=ksk=smg=0.25m39.5kg⋅9.81m/s2=1549.98 N/m39.5⋅dt2d2x=−1549.98⋅xdt2d2x+39.24⋅x=0The initial conditions are: x(0)=0.75 m,x′(0)=−1.25 m/s.
The negative sign in the last condition is a consequence of the fact that the mass is given an initial velocity in the negative, or upward, direction. Now ω2=39.24 or ω=6.26, so that the general solution of the differential equation is
x(t)=c1cos(6.26t)+c2sin(6.26t) Applying the initial conditions to x(t) and x′(t) gives
x(0)=c1=0.75
x′(t)=−0.75(6.26)sin(6.26t)+6.26c2cos(6.26t)
x′(0)=6.26c2=−1.25c2=6.261.25≈−0.20 Thus the equation of motion is
x(t)=0.75cos(6.26t)−0.20sin(6.26t)
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