Question

An object with mass 3.0 kg is executing simple harmonic motion, attached to a spring with spring constant k =210 N/m. When the object is 0.23 m from its equilibrium position, it is moving with a speed of 0.72 m/s.
(a) Calculate the amplitude of the motion.
m

(b) Calculate the maximum speed attained by the object.
m/s

Answer 1

F = m a = -kx

if x = A cos w t
v = dx/dt = -Aw sin w t
a = d^2x/dt^2 = -A w^2 cos w t = -w^2 x

so
-m w^2 x = -k x
w^2 = k/m
w^2 = 210/3 = 70
w = sqrt(70) = 8.37

when |x| = .23 , |v| = .72
.23 = A cos 8.37 t
.72 = -A (8.37) sin 8.37 t
so
cos 8.37 t = .23/A
sin 8.37 t = .086/A

cos^2 + sin^2 = 1
.0529 + .007399 = A^2
A^2 = .060299
A = .246 m
Vmax = Aw = .246(8.37) = 2.06 m/s

check my arithmetic !!!