# spring mass undamped motion

• Apr 5th 2011, 12:02 PM
duaneg37
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
• Apr 5th 2011, 01:25 PM
TheEmptySet
Quote:

Originally Posted by duaneg37
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
• Apr 5th 2011, 03:18 PM
duaneg37
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!
• Apr 5th 2011, 03:40 PM
TheEmptySet
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.
• Apr 5th 2011, 04:29 PM
topsquark
Quote:

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

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
• Apr 5th 2011, 04:35 PM
duaneg37
I think I've done them all wrong! Thanks a lot for you help!
• Apr 5th 2011, 07:26 PM
duaneg37
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?
• Apr 5th 2011, 07:40 PM
topsquark
Quote:

Originally Posted by duaneg37
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. (Nod)

-Dan