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Simple Harmonic Motion

  1. Jun 6, 2004 #1
    i'm stuck on two problems deal wih SHM

    1)A mass of .5 kg is attached to a vertical spring and slowly lowered to its equilibrium position; this stretches the spring by 10 cm. Then the mass is placed below in the position whre the stretch of the spring is 15 cm, and released. calculate the distance between the highest point that the moving mass reaches and the reference level corresponding to unstrained spring.

    ok, i have no clue. from the formulas that i have on SHM, i dont see how i can compute the highest point or any reference point.

    2)a horizontal spring on a frictionless table has a mass 0.1 kg attached to its end. spring constant is 0.5 N/m. the spring is stretched in a way that its elongation is 10 cm and then released, so the vibration begins. find teh velocity of the attached mass at the instant of time when the elongation of spring is 5 cm. find also the maximum volocity of the mass.

    i got the answer for the max velocity but not sure if its the right way to do it.
    ok, Vmax=w*A, where w is angular frequency of motion and A is amplitude

    so, w=2*PI*v=2PI/T=sqr(k/m

    to get T, i used this equation: T=2*pi/w=2*pi*sqrt(m/k

    so T=2*pi*sqrt(0.1 kg/0.5 N/m)=2.80

    so w=2*pi/2.80=2.237

    now, is this the correct formula to get Amplitude, a=-A*w^2
    anyway, i used it, to get A=0.5 N/m/2.237^2=.100

    so now i have A and w to get max velocity, Vmax=2.237*.100=.224 m/s???

    i have no idea how find the velocity fo the attached mass at the instant of time when elongation of is 5 cm

    now Final is tomorrow, so if you guys could show me the way to do these problems with the equations and all i'd really, really appreciate it
    thanks
     
  2. jcsd
  3. Jun 6, 2004 #2

    arildno

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    These types of exercises are most easily solved by conservation of mechanical energy.

    In general, for exercise 1, we have:
    [tex]\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}-mgx=C[/tex]
    where m is the mass, v the velocity, k the spring constant, x the elongation, g the acceleration of gravity and C an energy constant for the system.
    Note that I have chosen as reference level for the gravity potential to be the level at which the spring has 0 elongation.
    Positive x-direction is stretch direction, which is also the direction of the force of gravity.

    The equilibrium position is the position [tex]x_{0}[/tex] where the spring force and the force of gravity balances, i.e:
    [tex]mg-kx_{0}=0\to{k}=\frac{mg}{x_{0}}[/tex]

    This equation determines k.

    We are given that [tex]x_{0}=10[/tex] whereas we release the particle from rest in a strecthed position 5 length units further down.

    We are therefore able to determine the energy constant C:
    [tex]\frac{1}{2}m0^{2}+\frac{1}{2}k(15)^{2}-mg15=C[/tex]

    or:
    [tex]C=-\frac{15mg}{4}[/tex]

    You are now able to answer the question by remembering at the highest point, the velocity is also zero.

    For exercise 2, construct a similar energy conservation law, without the gravity potential.
     
  4. Jun 6, 2004 #3
    ok, i followed you for the most part, but for the hightest point if the velocity is also zero then what's gonna be the difference in the other variables.

    mass is the same, elongation, acceleration, gravity these dont change and constant is the same. so why won't it be c=-15mg/4 for the hights point either. please help
     
  5. Jun 6, 2004 #4

    Doc Al

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    To find (for problem 1) the highest point, set v = 0 in arildno's equation and solve for x:
    [tex]\frac{1}{2}kx^{2}-mgx=C[/tex]
    You already know m, k, and C, so solve this for x. It's a quadratic, with two solutions: the high point and the low point of the oscillation. (Note: I would recompute C in standard units, using meters instead of cm; I get C = -0.0375 mg.)

    Another way to solve it is to just realize that the mass will execute SHM about the equilibrium point, which is at x = - 10 cm. If you pull it to x = -15 cm, that's 5 cm away from equilibrium: it will oscillate between + 5 to - 5 cm about equilibrium. That's from -5 cm to -15 cm from the reference point.
     
  6. Jun 6, 2004 #5
    i'm working on it now
     
    Last edited by a moderator: Jun 6, 2004
  7. Jun 6, 2004 #6
    ok,
    (1/2)(49)x^2-(.5)(9,8)x+.184=0

    24.5x^2-4.9x+.184=0

    solved x by using the quadratic formula,

    4.9+/-Sqrt(4.9^2-(4)(24.5)(.184)/2(24.5)

    this gives me some really odd looking numbers.

    please help! Final exam is less than 18 hours away!!!
     
    Last edited by a moderator: Jun 6, 2004
  8. Jun 7, 2004 #7

    arildno

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    You get odd looking numbers because you introduce decimal representation far too early!
    NEVER EVER introduce "actual numbers" until you have simplified your "symbolic" equation maximally!!
    The ONLY exception to this rule, is when introducing "actual numbers" simplifies your equation.

    I will show how you should have done this:

    Let the initial position be:
    [tex]15=x_ {i}=x_{0}+5=x_{0}+\delta{x}[/tex]

    Hence, we have:
    [itex]C=\frac{1}{2}k(x_{i})^{2}-mg(x_{i})[/itex]

    Or we have, for the positions where the velocity is zero:
    [itex]\frac{1}{2}kx^{2}-\frac{1}{2}kx_{i}^{2}-mgx+mgx_{i}=0[/itex]
    This may be rewritten as:
    [itex](x-x_{i})(\frac{1}{2}k(x+x_{i})-mg)=0[/itex]
    Or the other root is given by:
    [itex]x=\frac{2mg}{k}-x_{i}[/itex]
    Remembering that [itex]k=\frac{mg}{x_{0}}[/itex] yields:
    [itex]x=2x_{0}-x_{i}=2x_{0}-(x_{0}+\delta{x})=x_{0}-\delta{x}[/itex]

    This shows, as Doc Al said, that the spring oscillates around the equilibrium position [itex]x_{0}[/itex]
    with amplitude [itex]\delta{x}[/itex]
     
    Last edited: Jun 7, 2004
  9. Jun 7, 2004 #8

    Doc Al

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    You get odd looking answers because you are making mistakes. But I certainly agree with arildno: don't plug in numbers until the last possible moment. For example, mg cancels from every term. (Unfortunately, I can't read arildno's latest response--some kind of format error. I'm sure he explains it all. :smile: )

    The quadratic I get is:
    1/2(mg/0.1)x^2 - mgx + (0.0375)mg = 0

    Simplify and solve this and you'll get the same simple answer as I state in my earlier post.
     
  10. Jun 7, 2004 #9

    arildno

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    Finally, I got it right :smile:
     
    Last edited: Jun 7, 2004
  11. Jun 7, 2004 #10
    i simplifed your equation and ended up with the same equation i had, 24.5x^2-4.9x+.184=0 Again if i solve for x this should give me the highest point and the reference point???the two vaules i get when i solved for x are .187 and .013............. whats the next step? sorry its that i'm not following to well i guess. i seem to be getting it. if you guys solved this problem, and i looked at how it was done i think i'll get it. sorry again its that i'm really short on time
     
    Last edited by a moderator: Jun 7, 2004
  12. Jun 7, 2004 #11

    arildno

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    Look at my previous post, and pay close attention to the derivation there
    (the formatting problems should be fixed by now)
     
  13. Jun 7, 2004 #12

    arildno

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    Be precise in identifying the steps you don't understand there!
     
  14. Jun 7, 2004 #13
    working on it
     
    Last edited by a moderator: Jun 7, 2004
  15. Jun 7, 2004 #14
    if mg cancel out from Doc Al's equation i should get 1/2/.1x^2-x+.0375=0----------------->25x^2-x+.0375=0
    solve for x??
     
  16. Jun 7, 2004 #15

    arildno

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    Again:
    DO NOT PUT IN NUMBERS BEFORE YOU NEED TO DO IT!!!!!
    Read post 7 closely.
     
  17. Jun 7, 2004 #16
    i dont get it! dag please help!!
     
  18. Jun 7, 2004 #17
    have mercy on my soul!
     
  19. Jun 7, 2004 #18
    its not a homework problem i swear, just please solve it for me
     
  20. Jun 7, 2004 #19

    arildno

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    1. The equation for the points where the velocity is zero is:
    [tex]\frac{1}{2}kx^{2}-\frac{1}{2}kx_{i}^{2}-mgx+mgx_{i}=0[/tex]
    (Definitions of the quantities are given in 7)
    2. This equation may be rewritten as:
    [tex](x-x_{i})(\frac{1}{2}k(x+x_{i})-mg)=0[/tex]
    This has 2 solutions:
    a) [tex]x=x_{i}[/tex]
    This is the initial position
    b)[tex]x=\frac{2mg}{k}-x_{i}[/tex]
    This is the maximal height position you're after.
    3)
    We have the relation: [tex]k=\frac{mg}{x_{0}}[/tex]
    Inserting this into 2b) yields:
    [tex]x=2x_{0}-x_{i}[/tex]
    4)
    Using the relation [tex]x_{i}=x_{0}+\delta{x}[/tex] we have:
    [tex]x=x_{0}-\delta{x}[/tex]
    That is, the spring oscillates with amplitude [tex]\delta{x}[/tex] about equilibrium position [tex]x_{0}[/tex]
    5)
    We know that:
    [tex]x_{0}=10[/tex] [tex]\delta{x}=5[/tex]
    Hence, we find that the top position have value: 10-5=5.
    The maximum height lies therefore 5 length units below the position of unstretched spring.

    Good luck!
     
  21. Jun 7, 2004 #20

    Doc Al

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    Not if you do it correctly! :smile: (1/2)/.1 = 5, not 25.

    Again, I agree with arildno's advice to not be in a hurry to plug in numbers. But if you do plug in numbers, you'd better do it correctly.
     
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