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Finding aplitude of free undamped motion

  1. Jul 12, 2009 #1
    I have a problem I'm trying to solve:

    A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 2ft/s. How many complete cycles will the mass have completed at the end of 4[tex]\pi[/tex] seconds?

    I understand that F=kx so I can say 32=k(2) therefore k=16. And a period can be found by T= 2[tex]\pi[/tex]/[tex]\omega[/tex] where [tex]\omega[/tex]=[tex]\sqrt{k/m}[/tex]. So in this case [tex]\omega[/tex]=4 and so the period, T, equals [tex]\pi[/tex]/2. What I can't figure out though is how to find the amplitude of motion. Any ideas? Thanks for the help.
     
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  3. Jul 12, 2009 #2

    djeitnstine

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    Since this is in the differential equations section I assume you need to solve a differential equation in order to get the information you need. Once you do so, the amplitude becomes obvious.

    Have you tried constructing an ODE yet? Since it is undamped the solution is quite simple.
     
  4. Jul 13, 2009 #3

    HallsofIvy

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    Presumably, you have a function of the form [itex]f(t)= A cos(\omega t)+ B sin(\omega t)[/itex]. There are two ways to find the amplitude:

    1) Find max and min values by differentiating. [itex]f'(t)= -A sin(\omega t)+ B cos(\omega t)= 0[/itex] gives [itex]B cos(\omega t)= A sin(\omega t)[/itex] so [itex]tan(\omega t)= A/B[/itex] [itex]t= arctan(A/B)/\omega[/itex]. Putting that back into the f(t) gives [itex]A cos(arctan(A/B))+ B sin(arctan(A/B))[/itex].

    You could use trig identities to change those cosine and sine functions into tangents but simpler is this: imagine a right triangle with opposite side A, near side B. By the Pythagorean theorem, the hypotenuse has length [itex]\sqrt{A^2+ B^2}[/itex]. Then the angle is arctan(A/B) and cosine of that angle is [itex]B/\sqrt{A^2+ B^2}[/itex] while the sine is [itex]A/\sqrt{A^2+ B^2}[/itex].

    2) Convert [itex]Acos(\omega t)+ Bsin(\omega t)[/itex] to a single function. sin(x+y)= sin(y)cos(x)+ cos(y)sin(x) so, taking [itex]x= \omega t[/itex], we just need to find "y" so that sin(y)= A and cos(y)= B. Of course, we can't do that is [itex]A^2+ B^2[/itex] is not equal to 1 but we can fix that by multiplying and dividing by [itex]\sqrt{A^2+ B^2}[/itex].
    [tex]A cos(\omega t)+ B sin(\omega t)= \sqrt{A^2+ B^2}\left(\frac{A}{\sqrt{A^2+ B^2}}cos(\omega t)+ \frac{B}{\sqrt{A^2+ B^2}}sin(\omega t)\right)[/tex]
    [tex]= \sqrt{A^2+ B^2}cos(\omega t+ y)[/tex]
    where y is such that [itex]sin(y)= A/\sqrt{A^2+ B^2}[/itex] and [itex]sin(y)= B/\sqrt{A^2+ B^2}[/itex].

    Of course, it is easy to look at that last formula and see what the amplitude is without need to find "y".
     
    Last edited: Mar 28, 2011
  5. Mar 28, 2011 #4
    My trig is a bit rusty.... How do you go from Acos(wt) +Bsin(wt) to sin(wt+y)
     
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