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Math Help - spring mass undamped motion

  1. #1
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    spring mass undamped motion

    A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.

    At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?

    my position function is x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)<br />

    I have tried to put it into the form  x(t)=Asin(\omega t+\phi)<br />

    and I am getting x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})<br />

    The time I am getting is \frac{\pi}{480}s initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks
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    Quote Originally Posted by duaneg37 View Post
    A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.

    At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?

    my position function is x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)<br />

    I have tried to put it into the form  x(t)=Asin(\omega t+\phi)<br />

    and I am getting x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})<br />

    The time I am getting is \frac{\pi}{480}s initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks
    First I don't think your ODE is correct.

    First by Hook's law we have

    F=kx \iff 10=k(.25) \iff k=40

    Now the ode is

    \displaystyle mx''=-kx \iff x''+\frac{1}{5}x=0

    So the solution should have the form

    \displaystyle x(t)=c_1\cos\left( \frac{t}{5}\right)+c_2\sin\left( \frac{t}{5}\right)

    Now use the initial conditions
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    I thought I had to convert the mass into slugs by W=mg, which gives the force as 320 slugs =k(\frac{1}{4}) for Hooke's law. This gives k=1280. Then I found m for the weight of 1.6  slugs >>> 1.6=m(32)  where m=.05 This gives  \omega ^{2}=25600. Is this incorrect?

    Shouldn't your equation be x(t)=c_{1}cos2t+c_{2}sin2t?

    I did it again without rounding off as much and got:

    x(t)=\frac{\sqrt{16393}}{384}sin(160t-1.547)

    the time I got:

    t=.026+\frac{n\pi}{160} sec., let     n=0,1,2,3,...

    Am I on the wrong track with this? Thanks a lot!
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    Weight is a force. Slugs are mass. So in hooks law you need a force and since lbs are a force you do not need to do any conversions.
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    Quote Originally Posted by TheEmptySet View Post
    \displaystyle mx''=-kx \iff x''+\frac{1}{5}x=0
    Actually \displaystyle \frac{k}{m} = \frac{40}{1.6} = 25, not 1/25.

    Thus
    x(t) = c_1~cos(5t) + c_2~sin(5t)

    -Dan
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    I think I've done them all wrong! Thanks a lot for you help!
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    My position function is x(t)=\frac{5}{12}sin(5t-.927)

    To find the time I set \frac{1}{2}=sin(5t-.927) I know the sine of \frac{\pi}{6} and \frac{5\pi}{6} will give \frac{1}{2}, so I get t=.29 and t=.709. Do I use both of these times to get my answer? I found the period to be T=\frac{2\pi}{5}. I said t=.29+\frac{2n\pi}{5} s where n=0,1,2,3,... and t=.709+\frac{2n\pi}{5} s where n=0,1,2,3,... as my answer. Am I doing this right?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duaneg37 View Post
    My position function is x(t)=\frac{5}{12}sin(5t-.927)

    To find the time I set \frac{1}{2}=sin(5t-.927) I know the sine of \frac{\pi}{6} and \frac{5\pi}{6} will give \frac{1}{2}, so I get t=.29 and t=.709. Do I use both of these times to get my answer? I found the period to be T=\frac{2\pi}{5}. I said t=.29+\frac{2n\pi}{5} s where n=0,1,2,3,... and t=.709+\frac{2n\pi}{5} s where n=0,1,2,3,... as my answer. Am I doing this right?
    Looks good to me.

    -Dan
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