1
$\begingroup$

I am self-studying through Mathematical Methods for Physics and Engineering and have come to a bit of a roadblock in chapter nine, on normal modes.

As pictured, the system is a uniform rod attached to a fixed point $P$ by way of a light string. The potential energy is given in the book as \begin{align} V &= Mlg\left[(1-\cos\theta_{1})+\frac{1}{2}(1-\cos\theta_{2})\right]\\ &\approx \frac{1}{4}Mlg\left(2\theta_{1}^{2}+\theta_{2}^{2}\right) \end{align}

My trouble is in transforming the first equation into the second; I have tried multiplying and dividing by the term in brackets but that didn't help. and I'm unsure of how to proceed.

Thank you!

$\endgroup$

1 Answer 1

0
$\begingroup$

Use the fact that $$ 1- \cos \theta = 2 \sin^2 \frac{\theta}{2} = 2 \left( \frac{\theta}{2} + O(\theta^2)\right)^2 = \frac{1}{2}\theta^2 + O(\theta^4) \textrm{ as } \theta \to 0 $$ So for small $\theta_1, \theta_2$ $$ (1-\cos \theta_1 ) + \frac{1}{2}(1-\cos \theta_2) \approx \frac{1}{2} \theta_1^2 + \frac{1}{4} \theta_2^2 $$

$\endgroup$
2
  • $\begingroup$ Thank you! Does that identity have a name (so I can reference it in the future?) $\endgroup$
    – Luciano
    Apr 12, 2017 at 1:10
  • $\begingroup$ Taylor series expansion of $\sin \theta$. $\endgroup$ Apr 12, 2017 at 1:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .