M3 SHM

Please could someone explain where the [x-(L/8)] comes from in the working for part b) in this question?

It's from the Jan 2014 IAL paper

Thanks!

Original post by PhyM23

Please could someone explain where the [x-(L/8)] comes from in the working for part b) in this question?

It's the extension from the equilibrium

x-e = x - \frac{\ell}{8}

which is the extension component in your tension for Hooke's Law.

Reply 2

φM23

Original post by Zacken

It's the extension from the equilibrium

x-e = x - \frac{\ell}{8}

which is the extension component in your tension for Hooke's Law.

Why would you want the extension from the equilibrium position? Wouldn't you want the extension of the spring from its natural length?

Reply 3

Zacken

Original post by PhyM23

Why would you want the extension from the equilibrium position? Wouldn't you want the extension of the spring from its natural length?

Nopes. Get used to this, it's going to be in every SHM problem ever. SHM works with the origin as equilibrium.

Reply 4

φM23

Original post by Zacken

Nopes. Get used to this, it's going to be in every SHM problem ever. SHM works with the origin as equilibrium.

I understand that SHM works with the origin as equilibrium, but if the spring is in equilibrium but compressed like in part a), then that means the extension from the equilibrium point would be 0 at that point, meaning (x-e)=0. But there has to be a compression force there doesn't there? So how can (x-e)=0 at that point?