# Block and Spring (Simple Harmonic Motion Problem)

## Homework Statement

At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the block next pass x = 0, the place where the spring is unstretched?

## Homework Equations

phi=phase angle
angular frequency= sqrt(k/m)=w
x(t)=Acos(wt+phi)

## The Attempt at a Solution

After working through the givens, I got my equation for the harmonic motion to be
x(t) = 0.66 cos (3.4987t - 2pi/3)
Where the angular frequency is sqrt(61.2/5)=3.4987, I got the amplitude from conservation of energy to be .6607 meters and the phase angle is -2pi/3.

The next step, if I am not mistaken, is to solve for when the spring is at its equilibrium position, or when x(t1)=0. Here are the steps for what I did...

0=0.6607 cos (3.4987t - 2pi/3)
0=cos(3.4987t - 2pi/3)
cos-1(0)=3.4987t - 2pi/3
pi/2= 3.4987t -2pi/3
7pi/6=3.4987t
t=1.0476 seconds

which, according to the homework website, is not correct! Can anyone see where I made my mistake? Thank you!!

I think probably since the motion is starting at t=0, then x(t) is better to be written as A cos(wt) , (im not really sure) .. but i think it is something like that ..

i have a question for you, why did you take phi = -2pi/3

ehild
Homework Helper
3.4987t - 2pi/3 is sooner -pi/2 than pi/2.

ehild

I figured out what I did wrong in this situation... cos-1(0) could be either pi/2 or -pi/2, and because the motion is coming to the end of a complete cycle I should have used -pi/2.

To answer your question, I determined the phase shift=phi by solving the position equation. I knew at t=0, x(t)= -.33 meters, so you just solve x(t) = A cos (wt + f) for f, and I got -2pi/3.