1. ## 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:

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

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

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

Am I on the right track?

2. 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:

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

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

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

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

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

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

This is incorrect. Please see Halls post #4 below to see the error in my logic.
Thanks again

3. 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:

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

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

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

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 $\displaystyle m=W/g=4/32$.

-Dan

@ VonNemo19 You should really put units on the numbers you are reporting. It's just a good habit.

4. Originally Posted by TheEmptySet
Yes the only thing I see that needs any attention is the term

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

$\displaystyle \displaystyle -2\bigg|\frac{dx}{dt}\bigg|$
No, $\displaystyle -2\frac{dx}{dt}$ 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!