A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 2.0 m/s. Determine the amplitude A of the motion.

Answer 1

x = A sin 2 pi f t
v = 2 pi f A cos 2 pi f t
max speed is when cos = 1 or - 1
vmax = 2 pi f A = 2

F = m a = -kx

a = -(2 pi f)^2 A sin 2 pi f t
= - (2 pi f)^2 x
so
.2 (-2pi f)^2 x = -120 x
so
2 pi f = sqrt (k/m) but you knew that :)
2 pi f = sqrt(120/.2) = 24.5
but
2 pi f A = 2
24.5 A = 2
A = .0816 or about 8 centimeters