# Simple Harmonic Motion of a spring

## Homework Statement

A hanging spring stretches by 35.0cm when an object of mass 450g is hung on it at rest. In this situation we define its position as x=0. The object is pulled down an additional 18.0cm and released from rest to oscillate without friction. What is its position x at a moment 84.4s later?

## Homework Equations

x(t)= Acos($$\omega$$t+$$\Phi$$)
$$\omega$$=$$\sqrt{\frac{k}{m}}$$

## The Attempt at a Solution

I know that I need to use these two equations, but I don't know k. So I was thinking that I need to use the information that a 450g weight stretches the spring 35.0cm to find the spring constant k. I just wasn't sure how to do that. I was thinking I should just multiply 450 by 35 and divide by 9.8 because it is hanging, but this didn't work for me.

rock.freak667
Homework Helper

## The Attempt at a Solution

I know that I need to use these two equations, but I don't know k. So I was thinking that I need to use the information that a 450g weight stretches the spring 35.0cm to find the spring constant k.

Yes, good. So the weight would be equal to the force of the spring. So kδ=mg meaning that k/m = g/δ.

ω=√(k/m)

Then you have x(t)= Acos(ωt+φ)

They tell you they stretch it 18 cm and then release, so what is x(0) and x'(0) ?