# Dimensional Analysis Question

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 $$A^{-1}$$

• Should be proportional to..........what? Anyway, this page is good, but looks worse than it is, give it a try physics.uoguelph.ca/tutorials/dimanaly
– user108787
Aug 25 '16 at 5:45
• It gives me multiple choice options after "Should be proportional to..." but that is the entirety of the question. Aug 25 '16 at 5:47
• Thank you for the link by the way I really appreciate it! I'm new to physics so I could really use the help @count_to_10 Aug 25 '16 at 5:48
• Sorry,wrong comments, for another page.
– user108787
Aug 25 '16 at 6:02

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:

$$\frac{1}{A} \sqrt{\frac{m}{k}}$$

• Why is it $-1 = 2\beta$ Where did you get -1 from? How exactly did you get $\gamma = - \beta$ was it by canceling $M^{\gamma} and m^{\gamma}$ and do you just ignore time and length when canceling out the masses? Aug 25 '16 at 7:33
• @Bazfred Time, mass and length are all independent fundamental units. Since the final answer must not contain either mass or length, both those units have to cancel out independently. Lastly, we want $L^\alpha \frac{M^\beta}{T^{2\beta} L^{2\beta}} M^\gamma$ to have units of time. The only way that will happen is if $-1=2\beta$. Aug 25 '16 at 7:41

Dividing by $k$ means that the powers of all its units get multiplied by minus one. So after the squareroot the power the quantity length will be left with $L^1$. This can only be cancelled by dividing by $A$.

Check the dimensional analysis of the following for the heat transfer between a flowing fluid and surface of the sphere

H=2kDP^-1+0.6DP^-0.5 G^0.5 (viscosity gradient mue)^-0.17 CP^0.33 K^0.67

• Welcome to the Physics Stack exchange! In order to obtain high interest and possibly good answers to the question, please stick to the habits of SE, i.e. using Mathjax for writing formulas and define the symbols you are using. Jul 1 '19 at 13:00