Answer to Question #194340 in Physics for Abdullah Latif

Question

a) A 2.00-kg block is placed on a frictionless surface. A spring with a force constant of k= 32.00 N/m is attached to the block, and the opposite end of the spring is attached to the wall. The spring can be compressed or extended. The equilibrium position is marked as x = 0.00 m. Work is done on the block, pulling it out to x=+0.02 m. The block is released from rest and oscillates between x = +0.02 m and x = -0.02 m. The period of the motion is 1.57 s. Determine the equations of motion.

Answer 1

$$\omega=\frac{2\pi}{T}=\frac{2\pi}{1.57}=4\frac{rad}{s}\\v=4(0.02)=0.08\frac{m}{s} \\a=4^2(0.02)=0.32\frac{m}{s^2}$$

$$x=(0.02)\cos(4.00t)\\v=(−0.8)\sin(4.00t)\\a=(−0.32)\cos(4.00t)$$