# Thread: spring mass undamped motion

1. ## 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 $x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
$

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

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

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

2. 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 $x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
$

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

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

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

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

Now the ode is

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

So the solution should have the form

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

Now use the initial conditions

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

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

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

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

the time I got:

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

Am I on the wrong track with this? Thanks a lot!

4. 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.

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

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

-Dan

6. I think I've done them all wrong! Thanks a lot for you help!

7. My position function is $x(t)=\frac{5}{12}sin(5t-.927)$

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

8. Originally Posted by duaneg37
My position function is $x(t)=\frac{5}{12}sin(5t-.927)$

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

-Dan

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### a mass weighting 32 pound stretch a spring 2 feet . the mass is initially released from a point 1 feet below the equilibrium position with a downward velocity of 2 ft/s . find the equation of motion

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