Year 12s: what will be the first way you research your future options? >>

M3 Elastic Springs

I have attached a question and it's worked solution below: When I worked out the problem I did not include the factor

2\lambda

(RHS of equation P.E. loss = E.E. gain in the numerator of the fraction). I don't understand why it is

2\lambda

instead of just

\lambda

since the EE. gain is the work done in stretching the spring, which is

\frac{\lambda x^2}{2l}

. I think it is because we have 'two spring segments' attached on either side of the particle, although I am not sure whether that is the correct justification.

Screen Shot 2017-12-26 at 6.41.31 PM.png94.3KB

Reply 1

cotkhd

You are right. In their working out, we can see that they are only taking into account the LENGTH of one spring, so if we didn't times it by two, we'd only have the E.P.E in one spring only. To get the EPE in 'both' springs, we just multiply the eqn by two.