# 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.

Related Introductory Physics Homework Help News on Phys.org
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) ?