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Simple Harmonic Motion, Force as a Function of Time

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A box of mass M is suspended by a spring of stiffness k. A small block of mass m is placed inside the box. If the system is pulled downward by a distance d and then released from rest:

    a.) find the force between the bottom of the box and the block as a function of time;
    b.) for what value of d does the block just begin to leave the bottom of the box at the top of the vertical osscilations?


    2. Relevant equations

    F = -kx


    3. The attempt at a solution

    I'm not exactly sure how to proceed with this.

    I started with Fnet = mg + Mg -kx

    I've rewritten it as [itex]\ddot{x} + \frac{k}{m + M}x = g[/itex]

    where [itex]\ddot{x}[/itex] is the acceleration with respect to time

    should I now just solve the differential equation? That should result in x as a function of time, and then take the second derivative of that and multiply it by the mass of the system?

    the part b question seems pretty simple, though. Just take Fnet = mg + Mg -kx, set it equal to zero, replace x with d, and solve for d.
     
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  3. Sep 12, 2012 #2

    Simon Bridge

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    You start with two free-body diagrams, one for the small mass, and one for the box.
    You could try doing the little mass part in the non-inertial frame of the box: it experiences a varying "gravity" depending on the acceleration of the system.


    I suspect you are trying to make one force equation do the work of two.

    But isn't there a shortcut?
    You know the equations of motion for SHM right ... x(t)=dcos(wt) right?
    You can find the frequency w from the gravity and the spring constant?
    You can find the equation for the acceleration of the box and so the pseudo-gravity experienced by the passenger?
     
    Last edited: Sep 12, 2012
  4. Sep 12, 2012 #3
    so I have the forces acting on the box as -kx, Mg and the force of the block on the box

    and then the forces on the block are mg and the force of the box on the block

    and then due to newton's third law, I can solve each of those for the reaction-pair force and set them equal to each other?

    FM = MaM = FgM - kx + Fblock
    Fm = mam = Fgm - Fbox

    I set the positive x-direction to be in the same direction as the force of gravity, so that makes the restoring force negative, the force of the block on the box positive, and the force of the box on the block negative.

    So then I get

    Fblock = MaM - FgM + kx
    Fbox = Fgm - mam

    then set them equal to each other and get

    MaM - FgM + kx = Fgm - mam

    now, aM and am should be equal to each other as long as d is such that the block never loses contact with the box, right? Would that be an appropriate simplification?
     
  5. Sep 12, 2012 #4
    the shortcut... I think there is, but I don't know how to get from the start to where the shortcut is...
     
  6. Sep 12, 2012 #5

    Simon Bridge

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    Lets see if I follow you: picking "down" as your positive x direction

    From the box you get: [itex]Mg-kx+F=M\ddot{x}[/itex] ...(1)
    (here F is the interaction force of the box with the block.)
    I've written F as a constant - is that sensible? Or does the force of the block on the box vary with time?

    From the block you get: [itex]mg-F=m\ddot{x}[/itex] ...(2)
    (since we require the block to never leave the floor of the box, we can expect the accelerations to be the same.)

    from (2) [itex]F=mg-m\ddot{x}[/itex] sub into (1) to get:
    [itex]Mg-kx+(mg-m\ddot{x})=M\ddot{x}[/itex] ... tidying up:

    [itex](M+m)g-kx = (M+m)\ddot{x}[/itex] ... (3)

    ...which is how far you've got.

    Eq.3 is the equation for SHM of a mass M+m on a spring of constant k.
    That's useful because you know the solution to this one, it has form: [itex]x(t)=A\cos(\omega t + \phi)[/itex]
    The formula for the frequency is also something you can look up: [tex]\omega=\sqrt{\frac{k}{m}}[/tex] (... in general - modify for your specific case.)

    From there you can find the equation for the acceleration as a function of time just by differentiating.

    The thing to consider is that the block experiences varying "gravity" which depends on the acceleration of the box. How? (Note: this reference frame is different from the one used to get eq(1) ... what's the difference?)

    Hopefully I've left you the important conceptual parts of the problem and given you enough of a leg-up with the math ;)
     
  7. Sep 12, 2012 #6
    see I'm not sure about the interaction force (F) being constant, because the part a. of the problem wants me to find the force between the bottom of the box and the block as a function of time. That would be the same thing as F. F would vary over time because it's dependent on the acceleration of the entire system.

    This is basically the same thing as an elevator accelerating, except the elevator's acceleration is changing as it goes up and down.

    The block is experiencing a varying "gravity" because the net force acting on it is changing over time and is dependent on its position.


    I think what I'm confused about is how you go from your Eq.3 to this: [itex]x(t)=A\cos(\omega t + \phi)[/itex]

    Going from there to differentiating and getting the acceleration function is something I know how to do, but won't that give me the acceleration of the system? Will I have to put that into Eq.2 and solve for F, which would be the force between the box and the block as a function of time?
     
  8. Sep 12, 2012 #7

    Simon Bridge

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    From equation (3) you have to solve the differential equation.
    It is second order and linear ... guess at solutions of form x(t)=exp(λt).
    But you can easily verify that the cosine equation is a solution to (3) by substituting it in.
    You should already have the cosine equation in your notes for your course - or, perhaps you have it as a sine wave?

    As for sorting out F: you can figure it out by examining a similar situation with a constant acceleration in a straight line ...

    If you suspect the F depends on t then modify the equation accordingly and see if it matters ;)

    And the last bit - right at the maximum height, if the block is just about to leave the floor of the box, then what is F?
    Another way of thinking about it is - what is the largest the downwards acceleration of the box can be without leaving the block behind?
     
    Last edited: Sep 12, 2012
  9. Sep 12, 2012 #8
    At the maximum height... then F would be the force due to gravity.

    We got the equation with a sin, yes, so differentiating it to find the acceleration and then substituting it into your Eq.2 would get me F = mg + mω2dsin(ωt)

    so at the start, t = 0 so the only force is mg... does this make sense? I don't think it does... because at t = 0 the box is being released, so there should be the force due to the spring acting... right?

    I'm sorry, I feel like I'm really lost in regards to some important factor in putting this all together :(
     
  10. Sep 12, 2012 #9

    Simon Bridge

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    F is given in eq(2) remember ... at other times, mg-F=ma ... so, at the maximum height, you are saying that F=mg: mg-mg=0=ma? How can this be??

    Take care - make sure you understand how this equation relates to the situation:

    at t=0, what is x?
    does this agree with the sine wave?

    That is right ... the full sine equation is [itex]x(t)=A\sin(\omega t + \phi)[/itex] so you still have to determine the phase factor.

    Take a look at mine: why did I use a cosine instead of a sine?
     
  11. Sep 12, 2012 #10
    aha, at t = 0, x = d... not x = 0.

    you chose cosine for the position function because at t = 0, the position function is at its maximum.

    at maximum height, then x = -d, so for the position function we have -d = dcos(ωt +θ)

    can't I just choose the phase factor to be zero? (I couldn't find phi in the side bar so I put theta).

    ω in this case is = √(k/(M+m))

    so we simplify our x(-d) equation to get -1 = cos(ωt) and ωt = π (that looks like pi, I guess)

    which makes sense

    so we can just take the derivative of the position function x(t) = dcos(ωt) and get

    x''(t) = -ω2dcos(ωt)

    and now, since we know that when ωt = π, the system is at its max height, we can put that in and see that the acceleration of the system is a = ω2d

    which makes sense, now that I realize it, because at the max height, the acceleration will be pointing straight down, in the positive direction in this case

    so if we substitute x''(t) into the Eq.2 we get

    mg-F=m[-ω2dcos(ωt)]

    and F = mg + m[ω2dcos(ωt)]

    now I see that when t = 0, F = mg + mω2d

    but... I'm not sure that makes sense. When the box is at its maximum displacement in the positive x direction, the acceleration should be at it's maximum and negative... right? Should I be substituting into Eq.1?
     
  12. Sep 12, 2012 #11

    Simon Bridge

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    you'll find the phi in the Ʃ menu when you use the advanced editor ... or you can just type the LaTeX in by hand like I do ;)

    Can you choose any old phase? What's the point of having it then?
    I see you noticed why I used cosine instead of sign - note: [tex]\cos(\omega t) = \sin(\omega t + \frac{\pi}{2})[/tex]... see what the phase does?

    At t=0 the box is at the bottom ... it is stationary and picking up speed upwards.

    You only need eq.2 for this ... you know F at the top ... [itex]\omega t = \pi[/itex] so x(t)=-d ... you have one final constraint to go.
     
  13. Sep 12, 2012 #12
    so if x(t) = -d when ωt = [itex]\pi[/itex], then t = [itex]\frac{\pi}{\omega}[/itex]

    ...so then what *is* the point of the phase? Shouldn't it be [itex]\omega t + \phi = \pi[/itex]?

    or rather, I see what the phase does, I just don't see how I should handle it in this case
     
  14. Sep 12, 2012 #13

    Simon Bridge

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    The generic equation is [itex]x(t)=A\sin(\omega t + \phi)[/itex]

    You don't normally start out with x(0)=0 though do we? The phase is used to set the distance from equilibrium when we start the stopwatch.

    If you start the stopwatch when the mass is passing through the equilibrium position and going downwards then [itex]\phi=0[/itex] ... if it is passing through equilibrium but going up, then [itex]\phi=\pi[/itex]. (This since you chose down to be the positive direction.)

    In this case the stopwatch was started when x(0)=d ... the amplitude of the motion was also d (this is the most common starting condition). This means that [itex]\phi=\frac{\pi}{2}[/itex] since, at the start:
    [itex]x(0)=d\sin(0+\frac{\pi}{2}) = d\sin(\frac{\pi}{2})=d[/itex]

    note: [itex]x(\frac{\pi}{\omega})= d\sin(\pi+\frac{\pi}{2})=d\sin(\frac{3\pi}{2})=-d[/itex] ... see below:

    It is easier just to write: [itex]x(t)=d\cos(\omega t)[/itex] and have done with it: it is the same function.

    When [itex]t=\frac{\pi}{\omega}[/itex] then [itex]x(\frac{\pi}{\omega})=d\cos(\pi)=-d[/itex] ... which is when the box is at the top see?

    I suspect you need to familiarise yourself with the sine and cosine functions - especially when the angles are in radians.
     
  15. Sep 13, 2012 #14
    okay, I see now. And yeah, I haven't spent a whole lot of time working with those two functions. It would probably be conducive to my studies if I did some practice with them.

    So now we have the position function of the system w.r.t. time, right?

    It's x = dcos(ωt)

    and then the acceleration function is

    [itex]\ddot{x} = -\omega^{2}dcos(\omega t)[/itex]

    and we just substitute that into eq.2 for a

    and get what I got earlier F = mg + m[ω2dcos(ωt)]

    when t = 0, F = mg + ω2d... that would mean that the force on the block from the box is positive, but that makes no sense
     
  16. Sep 13, 2012 #15

    Simon Bridge

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    Well you forgot an m ... to find out why F is positive here you need to have seen my fbd :)
    I had F pointing opposite to mg.

    How about at the other end?
     
  17. Sep 24, 2012 #16
    I'm confused about how to find acceleration. I understand that it is needed to determine the force of reaction by F(m+M)=mg+ma. Can anyone explain it? I've got the x(t)=dcos(ωt) and ω=√(k/(m+M)). Where do I go from there? Do I just find the second derivative of x(t)? And then -ω2dcos(ωt) is acceleration? so my force of reaction is mg+m(-ω2dcos(ωt))? That doesn't seem right.
     
  18. Sep 24, 2012 #17

    Simon Bridge

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    What you did in post #14 ... yes - this gives you an equation for acceleration vs time.

    But you want the force at a particular time.
    Reread your question: at what point in the motion do you need to know the acceleration?
    What is special about the acceleration at that point?
     
  19. Sep 24, 2012 #18
    also, thanks for your help Simon, I kind of dropped off there because it was getting late and I needed to sleep because I had class at 9AM >.>
     
  20. Sep 24, 2012 #19

    Simon Bridge

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