# Oscillations (SHM)

1. May 22, 2012

### roam

1. The problem statement, all variables and given/known data

I'm trying to solve this problem:

http://img824.imageshack.us/img824/3513/prob1r.jpg [Broken]

3. The attempt at a solution

I rearranged the equation T=2π√m/k to find the spring constant:

$k= \frac{m}{\left( \frac{T}{2 \pi} \right)^2} = \frac{70}{\left( \frac{6}{2 \pi} \right)^2}= 76.7628 \ N/m$

To find the original unstretched length I solve for y in F=ky and subtract it from 60 m:

F=-ky

$y= \frac{-F}{k} = \frac{-mg}{k} = \frac{-70 \times 9.81}{76.7628} = -8.957 \ m$

The spring is stretched by 8.94 meters from its equilibrium position so the original length is:

60-8.94=51.0543 m

Is this right? And how do I find the damping constant for the cord? What formula do I have to use? Any help is greatly appreciated.

Last edited by a moderator: May 6, 2017
2. May 27, 2012

### Simon Bridge

This is not SHM - this is damped harmonic motion... you will have been given the formula for that. You can also look it up.