Question #105114
A mass weighing 39 5. kg. stretches a spring 1/4m. At t = 0 , the mass is released from a point 3/4m below the equilibrium position with an upward velocity of 5/4 m/sec.Determine the function x(t) that describes the subsequent free motion.
1
Expert's answer
2020-03-29T08:06:20-0400

The spring mass equation for free motion is


md2xdt2=kxm{d^2x \over dt^2}=-kx

Hooke’s Law


mg=ksmg=ksk=mgs=39.5kg9.81m/s20.25m=1549.98 N/mk={mg \over s}={39.5kg\cdot 9.81 m/s^2 \over 0.25m}= 1549.98\ N/m39.5d2xdt2=1549.98x39.5\cdot{d^2x \over dt^2}=-1549.98\cdot xd2xdt2+39.24x=0{d^2x \over dt^2}+39.24\cdot x=0

The initial conditions are: x(0)=0.75 m,x(0)=1.25 m/s.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\omega^2=39.24 or ω=6.26,\omega=6.26, so that the general solution of the differential equation is


x(t)=c1cos(6.26t)+c2sin(6.26t)x(t)=c_1\cos(6.26t)+c_2\sin(6.26t)

Applying the initial conditions to x(t)x(t) and x(t)x'(t) gives


x(0)=c1=0.75x(0)=c_1=0.75

x(t)=0.75(6.26)sin(6.26t)+6.26c2cos(6.26t)x'(t)=-0.75(6.26)\sin(6.26t)+6.26c_2\cos(6.26t)

x(0)=6.26c2=1.25x'(0)=6.26c_2=-1.25c2=1.256.260.20c_2={1.25 \over6.26}\approx-0.20

 Thus the equation of motion is


x(t)=0.75cos(6.26t)0.20sin(6.26t)x(t)=0.75\cos(6.26t)-0.20\sin(6.26t)

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