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.

Question:

What is the maximum distance the spring will be stretched

The maximum distance A the spring will be stretched is  .

Explanation:

We note that for an ideal spring

F = ma = -k·x

That is m× = - k×x

If the equation above has the following initial conditions

x(0) = -x₀

= v₀

we have

Where:

A = Amplitude

∅ = Phase Angle

A and ∅ are such that

Therefore tan ∅ = and

Also, the amplitude A = Maximum displacement from equilibrium or the maximum distance the spring will be stretched is given by;

A = .

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

the answer is to check all of them! they all apply well at least i think so!