Answer to Question #104801 in Differential Equations for Suraj Singh

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/second. Determine the function x(t) that describes the subsequent free motion.

Answer 1

From Newton's second law $m\ddot{x}=-kx$ where $k$ is a spring stiffness, typing $\omega^2=\frac{k}{m}$ hence $\omega=\sqrt\frac{k}{m}$ . From equilibrium $mg=kl$ ,where $l=\frac{1}{4}m$ then $k=\frac{mg}{l}$ then $\omega=\sqrt\frac{g}{l}\approx6.26s^-1$

$x(t)=a\cos(\omega t-\alpha)$ ,$v(t)=x_t'=-a\omega\sin(\omega t-\alpha)$

$x_0(0)=a\cos\alpha$ ,$v_0(0)=a\omega\sin\alpha$ , since $x_0=0.75m$ , $v_0=1.25\frac{m}{s}$ hence $a=\sqrt{x_0^2+\frac{v_0^2}{\omega^2}}=0.775$ m

$\frac{v_0}{x_0}=\frac{a\omega \sin\alpha}{a\cos\alpha}=\omega\tan\alpha$ ,hence $\tan\alpha=\frac{v_0}{x_0 \omega}=0.27$, hence $\alpha=\arctan0.27\approx15^0\approx0.083\pi$ then

$x(t)=0.775\cos(6.26t+0.083)m$