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Oscillators, Energy, Force

  1. Feb 22, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    For a certain oscillator the net force on the body with mass m is given by fx = -cx^3.

    One-quarter of a period is the time for the body to move from x=0 to x=A. Calculate this time and hence the period.

    Express your answer in terms of the variables A, m, and c.

    2. Relevant equations

    Ux = (cx^4)/4 < Calculated from previous question part

    T=2 pi sqrt(m/k)

    3. The attempt at a solution

    I have got using the T equation, as far as:

    = 8 pi sqrt (m/something)

    but I am not quite sure what he something is.

    Any ideas?

    TFM
     
  2. jcsd
  3. Feb 22, 2008 #2

    G01

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    Can you please show your calculations and reasoning that led you to that equation for the period?
     
  4. Feb 22, 2008 #3

    TFM

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    I think the I made a mistake

    I did 1/4T = 2 pi sqrt (m/something)
    getting T = (2*4)pi sqrt(m/something)

    I don't think that 4 should have been in there. :blushing:

    Thus

    T = 2*pi sqrt(m/something)

    TFM
     
  5. Feb 23, 2008 #4

    TFM

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    Is the 'something', which is k, fx = -cx^3. since this is the force on the mass, but is it the restoring force?

    TFM
     
  6. Feb 23, 2008 #5

    TFM

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    Looking at the question in the book, I am using the wrong equation.

    In a hint in the book (But not on Mastering Physics), I have to use the energy equation

    E = 0.5mv^2 + 0.5kx^2,

    and insert the potential energy calculated in previous qauestion:

    (cx^4)/4

    I am not sure what part it replaces though...

    you then solve for v, replace v with dx/dt, seperate the variables with x on one side, t on the other, then intetgrate then use u = x/a

    Currently, I just need to find out where do I insert the potential Energy? does it replace the E at the beginning of the equation?

    TFM
     
  7. Feb 23, 2008 #6

    G01

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    No. E represents the total energy. You want to replace the potential energy term. You now know that E represents total energy and that 1/2mv^2 represents kinetic. There is only one other term in the equation correct?
     
  8. Feb 23, 2008 #7

    siddharth

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    No, that isn't right. If the amplitude of the oscillation is a, then the time period is simply,

    [tex] T = \int dt = 4 \int_0^a \frac{dx}{v} [/tex]

    where v is the velocity and as the question states, one-quarter of a period is time to move from 0 to a.

    Since the total energy T+V = E is constant, you have a relation between v and x, and so can integrate and find the time period. Can you take it from here?
     
  9. Feb 24, 2008 #8

    TFM

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    I got E = 0.5mv^2 + (cx^4)/4

    E - 0.5 mv^2 = (cx^4)/4

    I have solved for v

    v = sqrt((2((cx^4)/4 - E))/m)

    Replacing V with dx/dt

    dx/dt = sqrt((2((cx^4)/4 - E))/m)

    1/dt = sqrt((2((cx^4)/4 - E))/m)dx

    this seems a very complicated intergral, though? Is it correct so far?

    TFM
     
  10. Feb 24, 2008 #9
    Hey,

    I'm stuck on the same problem, this is one of the ways I have tried to solve it;

    [tex]E=(mv^2)/2+(kx^2)/2[/tex]

    Let the potential energy be equal to zero which means all the energy is kinetic, thus all the potential energy is now kinetic energy (assuming no other forces are acting) giving;

    [tex]kx^2/2=mv^2/2[/tex]

    letting [tex]kx^2/2=cx^4/4[/tex] gives,

    [tex]cx^4/4=mv^2/2[/tex] solving for v gives,

    [tex]v=\sqrt{cx^4/2m}[/tex] then taking the integral,

    [tex]T=4\int 1/v[/tex]

    [tex]T=4(-1/x)\sqrt{2m/c}[/tex] which is negative and time can't be negative! where have I gone wrong?

    I thought a way round it would be to square and square root the -1/x and combine it with the rest,

    [tex]T=\sqrt{2m/cx^2}[/tex] then placing the integral limits in would give,

    [tex]T=\sqrt{2m/cA^2}[/tex]

    Is this an ok thing to do, it seems like cheating to me.

    Thanks for any help.
     
  11. Feb 24, 2008 #10

    siddharth

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    First, have you seen the Latex tutorial for this forum? It's a really easy way to display equations.


    Right.

    Yeah, looks right. To get the dependence on A, m and c, change your variable of integration and make it dimensionless.

    This way, the integral will only be a numerical constant, and you'll have the dependence on A,m and c.
     
  12. Feb 24, 2008 #11

    TFM

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    No, I have never geard of Latex before. Is there a link to it?

    I take it I should change dx, but what do I change it into?

    TFM
     
  13. Feb 24, 2008 #12

    siddharth

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    See this thread.

    Here's a hint.

    When v=0, E = (ca^4)/4, right? But you know energy is conserved. So, substitute for E in terms of a. Then, you need to remove a from the limits in the integral. Then, what substitution is immediately obvious?
     
  14. Feb 24, 2008 #13

    TFM

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    I'm not sure if I have gone about this the right way:

    [tex] 0=\sqrt{\frac{2}{m}*(\frac{cx^4}{4}-\frac{ca^4}{4})} [/tex]

    ?

    TFM
     
  15. Feb 24, 2008 #14

    siddharth

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    No. How did you get that?

    [tex] E = \frac{mv^2}{2} + \frac{cx^4}{4} [/tex]

    Note that E is constant!

    when x=a, v=0, so that [tex]E=\frac{ca^4}{4}[/tex]

    Now use this in your equation for the time period. I think you should be able to get the answer from here.
     
  16. Feb 24, 2008 #15

    TFM

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    My time period equation was:

    [tex] T=2*\pi\*\sqrt{\frac{m}{k}} [/tex]

    Where does the E fit in?

    TFM
     
  17. Feb 24, 2008 #16

    siddharth

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    But, that is true only for a simple harmonic oscillator, where the potential is of the form kx^2!

    The whole point of the previous posts was to derive what the time period will be when it is not simple harmonic. In this case, the time period will depend on the amplitude.

    You were on the right track at Post #11. What went wrong?
     
  18. Feb 24, 2008 #17

    TFM

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    So what I need to do is:

    [tex] \frac{1}{dt} = \sqrt{\frac{2}{m}(\frac{cm^4}{4} - E)} dx[/tex]

    Which I can cancel down to:

    [tex] \frac{1}{dt} = \sqrt{\frac{cx^4}{2m} - \frac{2E}{m}} dx [/tex]

    Replacing E with

    [tex]\frac{ca^4}{4}[/tex]

    giving:

    [tex] \frac{1}{dt} = \sqrt{\frac{cx^4}{2m} - \frac{2(\frac{ca^4}{4})}{m}} dx [/tex]

    So far, so good?

    TFM
     
  19. Feb 24, 2008 #18

    TFM

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    I have now reduced it down to:

    [tex]\int \frac{1}{dt} = \int \sqrt{\frac{c(x^4 - a^4)}{2m}} dx[/tex]

    Is this Correct?

    TFM
     
  20. Feb 24, 2008 #19

    siddharth

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    Yes. Rewriting what you wrote,

    [tex] T = 4 \int_0^a \sqrt{\frac{2m}{c(x^4-a^4)}} dx = 4 \sqrt {\frac{2m}{ca^4}} \int_0^a \left(\sqrt{\frac{1}{(x/a)^4 - 1)}}\right) dx[/tex]

    Now, what is the obvious substitution to make, to reduce the intergal to just a number, and find the relation of T with a,m and c?
     
    Last edited: Feb 24, 2008
  21. Feb 24, 2008 #20

    TFM

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    Would an acceptable substitution be v = x/a, where v is a random letter, giving:

    [tex] T = 4\sqrt{\frac{2m}{ca^4}}\int^1_0 (\sqrt{\frac{1}{v^4 - 1}}) \frac{dv}{a} [/tex]

    TFM

    Edit: Forgot to change limits

    Using v = x/a, x = 0, a

    Limits are 0 and 1
     
    Last edited: Feb 24, 2008
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