Application of Mechanical Similarity

1. Nov 28, 2009

msbell1

Hi,

I am working on Mechanics by Landau and Lifgarbagez, and I have almost completed section 10 on mechanical similarity (in Volume 1, 3rd edition). However, I'm having some difficulty solving the problems at the end of the section. I have tried both of them, and I even found the correct answer to the first problem, but I feel like I didn't solve it the right way, or that there must be a more general way to solve it. So here's the problem.

Find the ratios of times in the same path for particles having different masses, but the same potential energy.

The answer is t'/t = sqrt(m'/m)

Here's how I solved it. First, by reading the question I thought hmmmmm....this sounds like 2 different masses on springs (the springs are the same) oscillating with the same amplitude. So then I approached the solution with this picture in mind.

For a mass on a spring, I know the period is given by T=2pi/w where w = sqrt(k/m)

so t'/t = T'/T = (2pi/sqrt(k/m'))/(2pi/sqrt(k/m)) = sqrt(m'/k)/sqrt(m/k) = sqrt(m'/m)

Ok, so I found the answer that I was supposed to find, but this surely wasn't the approach that I was supposed to take. The authors have not even covered oscillations yet in this book. Can someone please help me solve this problem the way the authors intended it to be solved? Thanks!

2. Nov 28, 2009

physicsworks

"The same path for the particle" means that $$\alpha \equiv 1$$ for coordinates:
$$\frac{l'}{l}=\alpha \equiv 1$$
Having differnt masses, the kinetic energy is multiplied not by $$\alpha^2/\beta^2}$$ (see your book) but $$\gamma \alpha^2 / \beta^2$$, where $$\gamma=m'/m$$ and $$\beta=t'/t$$.
To leave equations of motion unaltered, we must have
$$\gamma \frac{\alpha^2}{\beta^2} = \alpha^k$$
where the left side represents the kinetic energy factor and the right side represents the potential energy factor.
Substituting into this equation $$\alpha=1$$ one can have:
$$\gamma=\beta^2$$ or $$\beta = \sqrt{\gamma}$$
or, finally:
$$t'/t = \sqrt{m'/m}$$

I hope this will help you to solve the second problem in the paragraph also :)

Good luck!

3. Nov 30, 2009

msbell1

Thank you very much--that helped a lot!

4. Nov 30, 2009

msbell1

And now I will display my solution to the second question (actually it's very similar to the first one--thanks again, by the way).

$$\frac{l'}{l} = \alpha = 1$$(same path again)

This time the kinetic energy is just changed by $$\frac{\alpha^{2}}{\beta^{2}}$$ where $$\beta = \frac{t'}{t}$$

Since potential energies differ by a constant factor, we can write $$\frac{U'}{U} = \gamma$$

Finally, we have $$\frac{\alpha^{2}}{\beta^{2}} = \gamma\alpha^{k}$$

Since $$\alpha = 1$$, we are left with

$$\frac{1}{\beta^{2}} = \gamma$$

$$\beta = \frac{1}{\sqrt{\gamma}}$$

$$\beta = \frac{1}{\sqrt{U'/U}}$$

$$\beta = \sqrt{\frac{U}{U'}}$$

5. Dec 1, 2009

physicsworks

msbell1, sounds good! keep working on Landau & Lifgarbagez.