Question

A 2kg object is attached to a spring and undergoes simple harmonic motion. At t=0 the object starts from rest, 10 cm from the equilibrium position. If the force constant of the spring k is equal to 75 N/m, calculate
a. the max speed
b. the max acceleration of the object. Fine the velocity of the object at t=5s

Answer 1

10 cm = 0.1 m

x = .1 cos 2 pi t/T

v = -(2 pi/T) .1 sin 2 pi t/T

a = - (2 pi/T)^2 .1 cos 2 pi t/T = -(2 pi/T)^2 x

F = - k x

F = m a = - 2 (2 pi/T)^2 x
k= 8 pi^2/T^2 = 75

T ^2 = 8 pi^2/75
T = 1.03 s
2pi /T = 6.12
(2 pi/T)^2 = 37.5

so
if v = -(2 pi/T) .1 sin 2 pi t/T
then v max = (2 pi/T) .1
=.612
and
if a = - (2 pi/T)^2 .1 cos 2 pi t/T
then a max = .1(2 pi/T)^2
=3.75