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Amplitude vs Location when KE=PE?

  1. Dec 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A 0.50-kg object is attached to an ideal massless spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
    (a) What is the amplitude of vibration?
    (b) At what location are the kinetic energy and the potential energy of the system the same?
    Correct Answer: a) 0.24 m, b) 0.17 m
    I'm not sure how part A is different from part B.

    2. Relevant equations
    Part A:

    3. The attempt at a solution
    Part A:
    A=\sqrt{\dfrac{(.5)(1.5)^2}{20}}=0.24 \ \texttt{m}
    Part B:
    Same thing..?
    x=\sqrt{\dfrac{(.5)(1.5)^2}{20}}=0.24 \ \texttt{m}
  2. jcsd
  3. Dec 13, 2015 #2
    x = 0.24m being the amplitude means that the kinetic energy at that point is clearly zero. For part B, remember that the total energy remains the same.
  4. Dec 13, 2015 #3
    Hmm.. I still don't understand how this would get 0.17 m as an answer for part B.
  5. Dec 13, 2015 #4
    If the total energy remains the same throughout the motion, at the point where KE=PE, what is the expression of the PE/KE in terms of the total energy E?
  6. Dec 13, 2015 #5
    [tex]\dfrac{PE}{KE}=1 \ ?\\
    E = 2PE = 2KE \ ?
    Last edited: Dec 13, 2015
  7. Dec 13, 2015 #6
    Yes, and you already know the total energy, correct? Substitute in the expression for PE and you'll be able to solve for the position.
  8. Dec 13, 2015 #7
    Isn't it still 0.24 m?
    E = PE + KE\\
    E = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\
    2PE = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\
    2(\dfrac{1}{2}kA^2)= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\
    kA^2= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\
    kA^2-\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2\\
    0.24 \ \texttt{m} = x
  9. Dec 13, 2015 #8
    No, look through your steps once again. In particular, what does your 4th step say? That the total energy is 2 times of 1/2kA2? How is that possible? The concept being tested here is once again, that the total energy is conserved throughout the motion.
  10. Dec 13, 2015 #9
    Another hint for you though. Instead of thinking in terms of E = 2PE = 2KE, why not think of it in terms of KE = PE = 1/2E?

    Mathematically equivalent but I hope this makes it clearer with a shift of perspective
  11. Dec 13, 2015 #10
    Still don't see it. :headbang:
    The only way I got an answer that rounds to 0.17 m was by doing:
    E=\dfrac{1}{2}mv^2+\dfrac{1}{2}kA^2=1.125 \ \texttt{J}
    And then using that value here:
    0.167705 \ \texttt{m}=x\\
    0.17 \ \texttt{m} \approx x
    But, I am not sure if that is correct and why we are allowed to do [itex]\dfrac{E}{2}=PE+\dfrac{KE}{2}\\[/itex], let alone why we should use this step at all to get the correct answer.
    Last edited: Dec 13, 2015
  12. Dec 13, 2015 #11
    I think you need to see that TOTAL energy E =( PE at displacement x) + (KE at displacement x)
    When displacement x = 0,....PE =0 and KE is max
    When displacement x = A,... KE = 0 and PE is max
    so it is correct to say that E = PEmax
    or E = KEmax
  13. Dec 13, 2015 #12


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    Where did that come from?
    You know the total energy. How much PE is there when there's equal KE and PE? What extension does that give?
  14. Dec 13, 2015 #13
    Still not sure how I use this for the problem. :frown:
    [itex]\dfrac{E}{2}=PE[/itex]? But, what exactly am I supposed to do with that? When I substitute this into the equation, it still get 0.24 m as the answer. It seems like I don't understand something fundamental about this or I keep substituting in the wrong thing.
  15. Dec 13, 2015 #14


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    I see nowhere in this thread where you have done that correctly.
    In your post #7, there is an error going from your second line to the third. This path was not going to work anyway because at the end you still have a v in there, and you do not know what v is at the point of interest.
    Please try using [itex]\dfrac{E}{2}=PE[/itex] and post your steps.
  16. Dec 13, 2015 #15
    k=20, m=0.5, v=1.5, x=?\\
    \texttt{which leads to}\\
    x=0.24 \ \texttt{m}
    Where am I going wrong?
    Or if I just use [itex]\dfrac{E}{2}=PE[/itex] itself:
    0.24 \ \texttt{m}=x
    Same answer..
    Last edited: Dec 13, 2015
  17. Dec 13, 2015 #16


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    All your algebra steps are correct. It is only when you plug in the values and solve for x that you somehow make a mistake. You should be getting 0.24 m, for some reason you are off by a factor of 10. Did you use the values given in your first post?
    (By the way, there are actually two possible answers for x, 0.24 m and -0.24 m. I don't know why they choose to give only the positive answer)
  18. Dec 13, 2015 #17
    Sorry, it is 0.24 m. I just accidentally typed 2.4 m (fixed it in previous post now).

    But, I am trying to get 0.17 m for part B, not 0.24 m.
  19. Dec 13, 2015 #18
    In your first post you used max KE = max PE to calculate the amplitude.....great....you are more than 1/2 way there
    Max PE = 1/2kA^2.......... PE at any other value of displacement, x is = 1/2kx^2
    What value of x gives you 1/2 the max energy?...
  20. Dec 13, 2015 #19


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    Ah, sorry I had misread the first post.

    I see. The problem here is that in your solution, the v should be the speed when the object has PE = KE, but this speed is not 1.5 m/s (that's the maximum speed). Yours steps are correct, but using v=1/5 m/s in the final step is the mistake. When writing KE and PE, on must keep track where these are evaluated.

    What you do know is that E = KE + PE and that at the equilibrium position, where PE =0, the speed is 1.5 m/s You therefore have a value for the total energy.

    When you write E/2 = KE, this is correct but keep in mind that on the right side, the speed in 1/2 mv^2 is the speed at the position where KE=PE so this is an unknown. You can use that equation to find the speed at that point if you want, although it is not asked so you don't have to.

    It is better to write E = PE + KE as E = PE + E/2 at the point of interest, and then solve for x.
  21. Dec 13, 2015 #20
    Okay, hopefully, I got it this time.
    \text{At the equilibrium position, PE=0, so:}\\
    E=\dfrac{1}{2}mv^2=\dfrac{1}{2}(.5)(1.5)^2=.5625 \text{ J}\\
    \text{So, because } \dfrac{E}{2}=KE,\\
    0.17 \text{ m}\approx x
    Is this the correct way of doing part B?
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