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Period of small oscillations

  1. Nov 9, 2013 #1
    Q: http://gyazo.com/1ee7eee0134c25a23b4ad7a6972e1e46

    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. jcsd
  3. Nov 9, 2013 #2

    TSny

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    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.
     
  4. Nov 10, 2013 #3
    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
  5. Nov 10, 2013 #4
    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
  6. Nov 10, 2013 #5

    TSny

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    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.
     
  7. Nov 11, 2013 #6
    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
  8. Nov 11, 2013 #7

    TSny

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    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.
     
  9. Nov 12, 2013 #8
    thanks a lot for this - just one confusion - how do you know that ## E = v_0^2/2 ##?
     
  10. Nov 12, 2013 #9
    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?
     
  11. Nov 12, 2013 #10

    TSny

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    Evaluate the total energy at x = 0.
     
  12. Nov 12, 2013 #11

    TSny

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    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)##.
     
  13. Nov 12, 2013 #12
    thanks very much... this is my first course in physics so I'm slightly struggling, but got there in the end!
     
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