The period of oscillation of a nonlinear oscillator depends on the mass $m$, with dimensions of $M$; a restoring force constant $k$ with dimensions of $ML^2T^2$, and the amplitude $A$, with dimensions of $L$. Dimensional analysis shows that the period of oscillation should be proportional to
I'm confused on what the question is asking me to do. I know that $m = [M]$ and $k = ML^{-2}T^{-2}$ and that $A= [L]$
I got my answer to be $L(M/MTL)^{1/2}$ my answer was incorrect, but I am not quite sure why.
Which is $A\cdot(m/k)^{1/2}$
I know the answer, however I do not understand why $A^{-1}$
Best Answer
We expect that the period (units of time) should be proportional to some combination of $A$, $k$, and $m$. Let us try:
$$A^\alpha k^\beta m^\gamma$$
where $\alpha$, $\beta$, and $\gamma$ are constants to be determined. Substituting in the units:
$$A^\alpha k^\beta m^\gamma = L^\alpha \frac{M^\beta}{T^{2\beta} L^{2\beta}} M^\gamma $$
If this quantity is to have units of time, then we must have:
$$ -1 = 2\beta $$ or, therefore, $\beta = -1/2$.
Further, the masses have to cancel out. Thus:
$$\gamma = - \beta$$ Therefore, $\gamma=1/2$.
Also, the lengths have to cancel out:
$$ \alpha = 2\beta$$
which means $\alpha=-1$.
Thus, the only combination of $A$, $k$, and $m$ that will have units that match a period (time) is:
$$\frac{1}{A} \sqrt{\frac{m}{k}}$$