Answer to Question #300182 in Physics for KeKo

Question

(a) the displacement x, in m, from the equilibrium position of a particle moving with simple harmonic motion, is given by, x= 0.05 sin 6t, where t is the time, in s, measured from an instant when x= 0. (1) state the amplitude of the oscillations. (2) calculate the time period of the oscillations and the maximum acceleration of the particle.

(b) A mass hanging from a spring-suspended vertically is displaced a small amount and released. By considering the force on the mass at the instant when the mass is released, show that the motion is simple harmonic and derive an expression for the time period. Assume that the spring obeys Hooke's law.

Answer 1

Given:

$x= 0.05 \sin 6t$

(a) the amplitude of the oscillations

$$A=0.05\:\rm m$$

the period of oscillations

$$T=\frac{2\pi}{\omega}=\frac{2\pi}{6}=2.1\:\rm s$$

the maximum acceleration

$$a_{\max}=A\omega^2=0.05*6^2=1.8\:\rm m/s^2$$

(b)

$$\ddot{x}+\omega^2x=0$$

$$\omega=\sqrt{k/m}$$