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Thread: 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 $\displaystyle x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
    $

    I have tried to put it into the form $\displaystyle x(t)=Asin(\omega t+\phi)
    $

    and I am getting $\displaystyle x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})
    $

    The time I am getting is $\displaystyle \frac{\pi}{480}s$ 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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  2. #2
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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 $\displaystyle x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
    $

    I have tried to put it into the form $\displaystyle x(t)=Asin(\omega t+\phi)
    $

    and I am getting $\displaystyle x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})
    $

    The time I am getting is $\displaystyle \frac{\pi}{480}s$ 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

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

    Now the ode is

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

    So the solution should have the form

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

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

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

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

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

    the time I got:

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

    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
    $\displaystyle \displaystyle mx''=-kx \iff x''+\frac{1}{5}x=0$
    Actually $\displaystyle \displaystyle \frac{k}{m} = \frac{40}{1.6} = 25$, not 1/25.

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

    -Dan
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    I think I've done them all wrong! Thanks a lot for you help!
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  7. #7
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    My position function is $\displaystyle x(t)=\frac{5}{12}sin(5t-.927)$

    To find the time I set $\displaystyle \frac{1}{2}=sin(5t-.927) $ I know the sine of $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{5\pi}{6}$ will give $\displaystyle \frac{1}{2}$, so I get $\displaystyle t=.29 $and $\displaystyle t=.709$. Do I use both of these times to get my answer? I found the period to be $\displaystyle T=\frac{2\pi}{5}$. I said $\displaystyle t=.29+\frac{2n\pi}{5} $ s where $\displaystyle n=0,1,2,3,...$ and $\displaystyle t=.709+\frac{2n\pi}{5}$ s where $\displaystyle n=0,1,2,3,...$ as my answer. Am I doing this right?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duaneg37 View Post
    My position function is $\displaystyle x(t)=\frac{5}{12}sin(5t-.927)$

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

    -Dan
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