A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 Nim, If the oscillation is started by elongating the spring 0.1 m and giving the block a speed of 3 m/s then the amplitude of the oscillation is:

Answer 1

F = - 100 x = m d^x/dt^2
let x = A sin (w t - p)
dx/dt = w A cos (w t - p)
d^2 x/dt^2= - w^2 A sin (wt - p) = - w^2 x
so
-100 x = -m w^2 x
w^2 = 100/m
w = sqrt (100/m) but you knew that
at t = 0, x = 0.1 and dx/dt = 3
A sin (-p) = 0.1
sqrt (100/m) A cos(-p) = 3 but 100/m = 400
20 A cos -p = 3
so
A sin p = -0.1
A cos p=0.15
tan p = - 0.667 = -2/3
p = -33.7 degrees or -0.588 radians
so sin p = -0.555
A(-0.555) = -0.1
A = 0.18 meter

Answer 2

Check my arithmetic. I did that fast.