Spring mass problems

**VonNemo19** · March 22nd 2011, 11:45 AM

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?

**TheEmptySet** · March 22nd 2011, 11:49 AM

Originally Posted by VonNemo19

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

**topsquark** · March 22nd 2011, 03:11 PM

Originally Posted by VonNemo19

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.

**HallsofIvy** · March 23rd 2011, 10:13 AM

Originally Posted by TheEmptySet

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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