Answer to Question #105114 in Differential Equations for Abhishek

Question

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.

Answer 1

The spring mass equation for free motion is

$$m{d^2x \over dt^2}=-kx$$

Hooke’s Law

$$mg=ks$$ $$k={mg \over s}={39.5kg\cdot 9.81 m/s^2 \over 0.25m}= 1549.98\ N/m$$ $$39.5\cdot{d^2x \over dt^2}=-1549.98\cdot x$$ $${d^2x \over dt^2}+39.24\cdot x=0$$

The initial conditions are: $x(0)=0.75\ m, x'(0)=-1.25 \ m/s.$

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 $\omega^2=39.24$ or $\omega=6.26,$ so that the general solution of the differential equation is

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

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

$$x(0)=c_1=0.75$$

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

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

 Thus the equation of motion is

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