Question #194340

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.


1
Expert's answer
2021-05-17T17:34:49-0400
ω=2πT=2π1.57=4radsv=4(0.02)=0.08msa=42(0.02)=0.32ms2\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)x=(0.02)\cos(4.00t)\\v=(−0.8)\sin(4.00t)\\a=(−0.32)\cos(4.00t)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Abdullah Latif
18.05.21, 05:00

Just amazing

LATEST TUTORIALS
APPROVED BY CLIENTS