Order of anharmonicity of a simple pendulum

[Saptarshi Sarkar]

Homework Statement

The anharmonicity in the potential of a simple pendulum is of the order of

1. $\theta$
2. $\theta^2$
3. $\theta^3$
4. $\theta^4$

Relevant Equations

$V=mgl(1-cos\theta)$

I know that the potential of a simple pendulum is given by the above formula and that we can expand $cos\theta$ to get

$V=mgl\left(\frac{\theta^2}{2}-\frac{\theta^4}{24}+...\right )$

I am guessing that the answer is $\theta^4$, but I am not sure what "order" means here.

 

[PeroK]

That would be my guess too.

 

[haruspex]

It seems at least as valid to say that the standard harmonic equation is $\ddot x=-k^2x$, but the pendulum is $\ddot x=-k^2\sin(x)=-k^2x+O(x^3)$, so the anharmonicity is $O(x^3)$.

 

[kuruman]

It seems at least as valid to say that the standard harmonic equation is $\ddot x=-k^2x$, but the pendulum is $\ddot x=-k^2\sin(x)=-k^2x+O(x^3)$, so the anharmonicity is $O(x^3)$.

Yes, but that is the anharmonicity in the force not the potential which is what the question asks. The potential is symmetric about the equilibrium position and has no odd terms in its expansion.

 

[haruspex]

Yes, but that is the anharmonicity in the force not the potential which is what the question asks. The potential is symmetric about the equilibrium position and has no odd terms in its expansion.

Thanks, I missed that it specified potential.