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Simple Harmonic Motion - Potential energy

  1. Feb 7, 2014 #1
    For a mass on a spring (vertical set up) why is potential energy U defined as 1/2 kx^2? This is just the elastic potential energy. Shouldn't it be U = 1/2 kx^2 + mgh? Both the elastic AND potential energy? Also, for a simple pendulum at a very low amplitude, the potential energy is all gravitational, right? By the way, if U = 1/2 kx^2 + mgh, then the relationship ω = √(k/m) becomes invalid, since U = 1/2 mω^2 x^2 is always valid and equating this to both elastic and gravitational potential energy gives us a different expression for ω.
     
  2. jcsd
  3. Feb 7, 2014 #2
    1/2 kx^2 is the only term that is not constant in the potential energy.
    The gravitational potential energy can add a constant to it. As you can always do for potential energy.
    Try to calculate is with gravity included and you will see it.
     
  4. Feb 7, 2014 #3

    olivermsun

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    Hint: Define x so that x=0 right where the system is at "rest." (This will not be the same position as the rest position in the absence of gravity.)
     
  5. Feb 7, 2014 #4

    russ_watters

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    Consider that if your rotate the system horizontal (and put it on a frictionless surface), the equations still work. Gravity only determines the equilibrium position, it doesn't determine the dynamics.
     
  6. Feb 7, 2014 #5

    I know it doesn't determine the dynamics, but this is not my doubt. My doubt is, 1/2 m w^2 x^2 must always equal 1/2 kx^2 yet this is not the case here, because there is gravitational potential energy varying as the mass oscillates up and down. The conservation of energy holds whether or not gravity affects the dynamics of the mass on the spring.
     
  7. Feb 7, 2014 #6

    How is gravitational potential energy constant? The mass is moving up and down. And K_A + U_totA = K_B + U_totB always holds as mechanical energy is conserved, where U_tot = U_elastic + U_grav. Can you elaborate more please?
     
  8. Feb 7, 2014 #7

    jtbell

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    First suppose there is no gravity, and we have the spring oriented vertically with one end fixed. Let x = 0 be the position of the mass when it is in equilibrium. Then
    $$U_{spring} = \frac{1}{2}kx^2$$
    Next, remove the spring for a moment, and "turn on" gravity. Define the gravitational potential energy such that it is zero when the mass is at the same x = 0 position as defined above. Then
    $$U_{grav} = mgx$$

    Under the influence of both the spring and gravity, we therefore have
    $$U = \frac{1}{2}kx^2 + mgx$$
    In this situation the mass has a new equilibrium position, where ##\vec F_{spring} + \vec F_{grav} = 0##.

    Exercise 1: Find the new equilibrium position... how far is it below the old equilibrium position?

    Exercise 2: Define a new vertical position variable, z, such that z = 0 at the new equilibrium position. Find an equation that relates z and x.

    Exercise 3: Change the variable in the equation for U above, from x to z, by making an appropriate substitution. What do you get?
     
  9. Feb 7, 2014 #8

    olivermsun

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    The point is that when you dangle the mass under its own weight, the equilibrium position moves from the point determined by the rest length of the spring (call it x = 0) to some new lower position (x = c < 0) where F = -kc = -mg. That is, the new rest position is exactly where the stretched spring exerts a restoring force that exactly opposes gravity.

    Due to the linearity of the governing equation, you can analyze the system as:

    F = -kx + mg = -kx + kc = -k(x-c) .

    So if you consider this as a system in the new variable x' = x - c, then you see that everything behaves exactly as before, except now the center of oscillation is around x' = 0 (where x = c).

    (Never mind, jtbell already posted the solution...)
     
    Last edited: Feb 7, 2014
  10. Feb 7, 2014 #9
    I did not say that the gravitational potential energy is constant, did I?
    I said that considering the gravitational potential energy results in adding a constant to the expression of the energy
    After you do the work, as suggested already by jtbel.
    The best way to understand is to try yourself and do the calculations.
     
  11. Feb 7, 2014 #10
    I got lost at Ex 2, can you please give me more hints?
     
  12. Feb 7, 2014 #11
    Do you mean to say that now the total potential energy is 1/2 k x' ^2 ?
     
  13. Feb 7, 2014 #12

    UltrafastPED

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  14. Feb 7, 2014 #13
    Was i supposed to get U = 1/2 kz^2? I ended up with a slightly more complicated answer.
     
  15. Feb 7, 2014 #14
    Apparently I made a wrong substitution in the equation U = mgx + 1/2 kx^2. Can you please tell me where I went wrong?
     
  16. Feb 7, 2014 #15

    jtbell

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    Correct!

    Show us what you did and someone will probably be able to find your error. When you do the substitution you should get something a bit messy, but most of it should cancel out, leaving only the 1/2 kz^2.
     
  17. Feb 7, 2014 #16

    jtbell

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    Here's a diagram which might help. I don't know what you got for Ex 1, so I'm calling it Δ. For any vertical position, what is the relationship between z and x?

    attachment.php?attachmentid=66390&stc=1&d=1391814424.gif
     

    Attached Files:

  18. Feb 8, 2014 #17
    http://postimg.org/image/68566ta6f/

    This is my working, where did I go wrong?

    U = 1/2 kx^2 + mgx, and x = z + c, and mg = kc
    I eventually got U = 1/2 kz^2 + 2kzc + 3/2 kc^2
     
  19. Feb 8, 2014 #18
    When I tried to modify what I did by using mg = -kc instead, i ended up with U = 1/2k(z^2 - c^2)
    Apparently I have a misconception regarding the signs. The thing is, throughout my mechanics courses, If I have an equilibrium problem in which, say, a mass is on a spring, I would equate T to mg where T is the tension and mg is the force of gravity, but I never did something like T = -mg, or anything of that sort. Where exactly am I going wrong?
     
  20. Feb 8, 2014 #19
    It's a matter of choosing the positive direction.
    You wrote the energy with +mgx for gravitational PE.
    x is measured from the position of equilibrium without gravity, right? Where the spring is not stretched.
    When the body goes down the gravitational PE must decrease. This means that x must be negative.
    So you implicitly assumed that negative is down and positive is up and your "x" is negative.
    This is why the condition of equilibrium is
    mg=-kx.

    You can avoid this by
    Writing the gravitational PE term as -mgx.
     
  21. Feb 8, 2014 #20
    But that still doesn't give U = 1/2 kz^2
     
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