# Period of small oscillations

1. Nov 9, 2013

### synkk

part a)

I have drawn the graph and calculated $V'(x) = \dfrac{3\lambda x^2 (x^4 + a^4) - \lambda x^3(4x^3)}{(x^4+a^4)^2} = 0$ and found using the graph that the value of x when the particle is in a stable equilibrium is $x= -3^{\frac{1}{4}}a$ but I'm not sure how to find the small period of oscillations, I know the formula is $\dfrac{2\pi}{w} = 2\pi \sqrt{\dfrac{m}{V''(X)}}$ but working out V''(x) is fine, but subbing in the value of x is a lot of algebra to do without a calculator and I have a feeling there is an easier way which I am missing

2. Nov 9, 2013

### TSny

I think your approach is good. I don't see an easier way. If you've already found an expression for V''(x), then it's not too bad to substitute the equilibrium value for x and simplify.

Note that the first derivative V'(x) has the form f(x)/g(x). At the equilibrium value xo, you have f(xo) = 0.

Then V''(xo) = f$\;$'(xo)/g(xo) - f(xo)g'(xo))/g(xo)^2.

The second term on the right is zero because f(xo) = 0.

3. Nov 10, 2013

### synkk

Thanks, that helped make it easier,

After plugging it all in I ended up with the period of small oscillations to be $2\pi \sqrt{\dfrac{4a^3}{3^{\frac{5}{4}}\lambda}}$ which seems strange, doesn't look "nice" and usually in my homework problems solutions usually look tidy.

Last edited: Nov 10, 2013
4. Nov 10, 2013

### synkk

Moving on to the next part of the question, I need to show that if x = a it implies that $v_0 > \frac{\lambda}{a}$ I done this using the energy equation $E = \frac{1}{2}v_0^2 + V(X)$ and subbing in x = a I get $v_0^2 + \frac{\lambda}{a} = 2E$ as E > 0 and 2E>0 then $v_0^2 > \frac{\lambda}{a}$ I'm having troubles proving it the other way round however,

i.e. if $v_0^2 > \frac{\lambda}{a}$ then x = a is it as simple to say that as $E > \frac{\lambda}{a}$ and $E = \frac{1}{2}v_0^2 + V(x)$ then V(x) must equal $\frac{\lambda}{2a}$ in order for E to be $> \frac{\lambda}{a}$ i.e. x = a ----------- NVM this is incorrect, using $E > \frac{\lambda}{a}$ is a result I got from assuming x = a :|

Also, generally I'm confused about the question, it states that if the particle in the positive x direction then show it will go through a, but if it goes through a then wouldn't the particle keep on moving to + infinity? If so, then how could it possible go back to -a

Last edited: Nov 10, 2013
5. Nov 10, 2013

### TSny

If $v_o^2 > \lambda/a$ then $E > \lambda/(2a)$. The only way the particle will not make it to $x = a$ is if the kinetic energy goes to zero at some value of $x$, say $x_{\rm max}$, such that $x_{\rm max} < a$. Think about what that means for the potential energy at $x_{\rm max}$ and use the sketch of your graph to argue that there is no such $x_{\rm max}$ that is less than $a$.

For the last question, note that it is possible for the particle to make it past $x = a$ without moving on to + infinity. Again, your graph might be helpful.

6. Nov 11, 2013

### synkk

ok, thanks for the tips, the reason I had troubles with believing it went back down is because I thought 3^(1/4)a was < a, lol.

Anyway, here is my attempt for proving the "only if":
if $v_0^2 > \frac{\lambda}{a}$ then as $V(x) > 0$ for all positive x then $E > \frac{\lambda}{2a}$

assume $x \not= a$ then $\exists x_{m}$ s.t. $x_m < a$

this implies that $V(x_m) = E$ as the particle stops moving - but $as V(x_m) < V(a) = \frac{\lambda}{2a}$ this implies that $E < \frac{\lambda}{2a}$ which is a contradiction hence x must equal a

is this correct?

for the last part:

if the particle goes to x = -a then the energy must be below the maximum point of the graph, i.e. $E < 3^{1/4}a \Rightarrow \frac{1}{2} v_0^2 + V(-a) < 3^{1/4}a \Rightarrow v_0^2 < 2\times3^{1/4}a - \frac{\lambda}{a}$

and $v_0^2 > \frac{-\lambda}{a}$ so $\frac{-\lambda}{a} < v_0^2 < 2\times3^{1/4}a - \frac{\lambda}{a}$

Last edited: Nov 11, 2013
7. Nov 11, 2013

### TSny

There is no need for the phrase "V(x) > 0 for all positive x". At x = 0 you can show that $E > \frac{\lambda}{2a}$ and by conservation of energy, $E > \frac{\lambda}{2a}$ for all values of x that the particle makes it to.

Yes, essentially. I had suggested that you could approach the problem by assuming that the particle does not make it past $x = a$ and show that this leads to a contradiction. Thus, if the particle does not make it to $x = a$ it must momentarily come to rest at some $x = x_m$ such that $x_m < a$. Since the graph of V(x) is monatonically increasing, $V(x_m) < V(a) = \frac{\lambda}{2a}$. Therefore, at $x = x_m$, $E = V(x_m) < \frac{\lambda}{2a}$. This contradicts the fact that we know the particle starts out with $E > \frac{\lambda}{2a}$. I think this is pretty much the same as your argument.

You just need to make sure the particle does not make it over the hill at $x = 3^{1/4}a$. So, argue that E must be less than $V(3^{1/4}a)$. You know that $E = v_0^2/2$. So, this should give you the restriction on $v_0^2$. Note that $v_0^2$ is certainly positive, so the lower limit of $v_0^2$ will be 0.

8. Nov 12, 2013

### synkk

thanks a lot for this - just one confusion - how do you know that $E = v_0^2/2$?

9. Nov 12, 2013

### synkk

I was thinking as $v_0^2/2 = E - V(x)$ as $E <V(3^{1/4}a)$ then $E - V(X) < \dfrac{3^{3/4}\lambda}{a}$ so $0 < v_0^2 < \dfrac{3^{3/4}\lambda}{2a}$ is my logic here correct?

10. Nov 12, 2013

### TSny

Evaluate the total energy at x = 0.

11. Nov 12, 2013

### TSny

The first equation should be $v_0^2/2 = E$. Since the graph of $V(x)$ has only one maximum (at $x = 3^{1/4}a$), the particle will stop, turn around, and make it to $x = -a$ as long as $E < V(3^{1/4}a)$.

12. Nov 12, 2013

### synkk

thanks very much... this is my first course in physics so I'm slightly struggling, but got there in the end!