Question

Consider an ideal spring that has an unstretched length ℓ and a spring constant k. Suppose the spring is attached to a mass m that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x_0 from equilibrium and then released with an initial speed v_0 toward the equilibrium position.

What is the maximum distance the spring will be stretched?

I tried conservation of energy but v_0 can't be used in the solution so I'm at a lost as to what to do.

Answer 1

x = a sin w t + b cos wt
v = a w cos wt - b w sin w t

omega = w = sqrt(k/m)
at t = 0
x = -xo
v = dx/dt = vo

- xo = b
+ vo = a w or a = vo/w

so x = (vo/w) sin w t - xo cos wt

now I could say
magnitude (or amplitude) = sqrt (vo^2/w^2 + xo^2) which is max motion from equilibrium

= sqrt (vo^2 m/k + xo^2)

Answer 2

but how to I eliminate v_0?

Answer 3

never mind I typed it wrong. Thanks!