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Math Help - Spring mass problems

  1. #1
    No one in Particular VonNemo19's Avatar
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    Spring mass problems

    Hi. So...

    Suppose that an object weighing 4lbs stretches a spring 2 inches. The object is displaced an additional six inches below the equilibrium position and then released. The object is in a medium that exerts a damping force nemuerically equal to 2 times the instantaneous velocity. Determine the equation of motion.

    So, Here I know that since F=ks by Hooke's Law, we can determine k as follows:

    4=k(\frac{1}{6})\Rightarrow{k}=24.

    Now, since from newton's second law we know that F=ma, where m=W/g=8/32=1/4 and a=\frac{d^2x}{dt^2} and we can equate this to the sum of the restoring force of the spring and the resistant force of the medium so that

    \frac{1}{4}\frac{d^2x}{dt^2}=-24x-2\frac{dx}{dt}

    Am I on the right track?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by VonNemo19 View Post
    Hi. So...

    Suppose that an object weighing 4lbs stretches a spring 2 inches. The object is displaced an additional six inches below the equilibrium position and then released. The object is in a medium that exerts a damping force nemuerically equal to 2 times the instantaneous velocity. Determine the equation of motion.

    So, Here I know that since F=ks by Hooke's Law, we can determine k as follows:

    4=k(\frac{1}{6})\Rightarrow{k}=24.

    Now, since from newton's second law we know that F=ma, where m=W/g=8/32=1/4 and a=\frac{d^2x}{dt^2} and we can equate this to the sum of the restoring force of the spring and the resistant force of the medium so that

    \frac{1}{4}\frac{d^2x}{dt^2}=-24x-2\frac{dx}{dt}

    Am I on the right track?
    Yes the only thing I see that needs any attention is the term

    \displaystyle -2\frac{dx}{dt} we really need absolute values because it causes resistance in both directions (both up and down) so

    \displaystyle -2\bigg|\frac{dx}{dt}\bigg|

    This is incorrect. Please see Halls post #4 below to see the error in my logic.
    Thanks again
    Last edited by TheEmptySet; March 23rd 2011 at 11:26 AM. Reason: explained above!
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Hi. So...

    Suppose that an object weighing 4lbs stretches a spring 2 inches. The object is displaced an additional six inches below the equilibrium position and then released. The object is in a medium that exerts a damping force nemuerically equal to 2 times the instantaneous velocity. Determine the equation of motion.

    So, Here I know that since F=ks by Hooke's Law, we can determine k as follows:

    4=k(\frac{1}{6})\Rightarrow{k}=24.

    Now, since from newton's second law we know that F=ma, where m=W/g=8/32=1/4 and a=\frac{d^2x}{dt^2} and we can equate this to the sum of the restoring force of the spring and the resistant force of the medium so that

    \frac{1}{4}\frac{d^2x}{dt^2}=-24x-2\frac{dx}{dt}

    Am I on the right track?
    Two corrections:
    1) Edit: Okay, that one was right. (2 in = 1/6 ft.)

    2) The weight is given as 4 lb, not 8 lb. So m=W/g=4/32.

    -Dan

    @ VonNemo19 You should really put units on the numbers you are reporting. It's just a good habit.
    Last edited by topsquark; March 22nd 2011 at 05:06 PM.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Yes the only thing I see that needs any attention is the term

    \displaystyle -2\frac{dx}{dt} we really need absolute values because it causes resistance in both directions (both up and down) so

    \displaystyle -2\bigg|\frac{dx}{dt}\bigg|
    No, -2\frac{dx}{dt} precisely because "it causes resistance in both directions (both up and down)", if the motion is upward, dx/dt> 0 so that -2(dx/dt) is negative (opposite the motion) and if the motion is downward, dx/dt< 0 so that -2(dx/dt) is positive, again opposite the motion. If you used -2|dx/dt|, when the motion was downward, the resistance force would also be dowhward!
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