Question

A 150 gram mass is attached to a horizontally aligned spring on a frictionless surface. A force of 10 Newtons will stretch the spring 25 centimeters. If the spring is compressed to 15 centi¬me¬ters and then released, calculate: the spring constant; the frequency and period of the system; and the position and speed of the mass one minute after it is released.

Answer 1

mass m = 150 g = 0.15 kg

Force F = 10 N

Stretchness X = 25 cm = 0.25 m

Spring constant K = F / X

= 10 / 0.25

= 40 N/m

Angular frequency w = [K/m]

= [40/0.15]

= 16.32 rad/s

Frequency f= w/2(pi)

= 16.32 / 2(3.1415)

= 2.598 Hz

Period of the system T = 1/ f

= 0.3847 s

position x(t) = (15 cm)cos wt

= (0.15 m) cos 16.32 t

time t = 1 min = 60 s

x(60) = 0.15 cos (16.32x60)

= 0.15 cos(979.2 rad )

= 0.15 (0.5595 )

= 0.0839 m

velocity v(t) = dx(t)/dt

= (0.15)(16.32) [-sin(16.32 t]