Answer to Question #325726 in Physics for Claire

Question

A 0.25-kg mass at the end of a spring oscillates 2.2 times per second with an amplitude of 0.15 m. Determine (a) the speed when it passes the equilibrium point, (b) the speed when it is 0.10 m from equilibrium, (c) the total energy of the system, and (d) the equation describing the motion of the mass, assuming that at t=0, x was a maximum.

Answer 1

The angular frequency of oscillation

$\omega=2\pi f\\ =2*3.14*2.2=13.8\:\rm rad/s$

(a)

$v_{\max}=\omega A\\ =13.8*0.15=2.1\;\rm m/s$

(b)

$v=\sqrt{v_{\max}^2-\omega^2x^2}\\ =\sqrt{2.1^2-13.8^2*0.10^2}=1.6\:\rm m/s$

(c)

$E=\frac{m\omega^2A^2}{2}\\ =\frac{0.25*13.8^2*0.15^2}{2}=0.54\:\rm J$

(d)

$x=0.15\cos(13.8t)$