# Thread: Application DE (dampened motion)

1. ## Application DE (dampened motion)

I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

then it ask: intaially the spring starts out at 1 meter below equilbrium

so i put that x(0) = 1

then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?

2. Originally Posted by neven87
I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

then it ask: intaially the spring starts out at 1 meter below equilbrium

so i put that x(0) = 1

then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?
First of all, what is "a 10 times dampening"?

The general equation is
$x^{\prime \prime} + \beta x^{\prime} + \omega_0^2 = 0$

$\beta$ is the damping constant. Are you trying to say that $\beta = 10~N/ms$?

-Dan

3. Originally Posted by neven87
I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

then it ask: intaially the spring starts out at 1 meter below equilbrium

so i put that x(0) = 1

then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?
Originally Posted by topsquark
First of all, what is "a 10 times dampening"?

The general equation is
$x^{\prime \prime} + \beta x^{\prime} + \omega_0^2 = 0$

$\beta$ is the damping constant. Are you trying to say that $\beta = 10~N/ms$?

-Dan
I'm going to assume my previous post is correct since it gives the same solution to the harmonic motion differential equation as the one you posted.

There is a second initial condition:
$x^{\prime}(0) = 0~m/s$
since the mass is released at 1 m.

-Dan

4. yes, you are correct, sorry the book only calls it beta...

yes, but the book's answer is two parted and with the derviatve i still get no where when i sub x(0) = 0 in

the answer is x(T)= 4/3e^-2t - 1/3e^-8t for the distance

and -2/3e^-2t + 5/3 e^-8t for the velocity.

5. Originally Posted by neven87
I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

x(0) = 1
Originally Posted by topsquark
There is a second initial condition:
$x^{\prime}(0) = 0~m/s$
since the mass is released at 1 m.

-Dan
So we have that
$1 = C_1 + C_2$

Now,
$x^{\prime}(t) = -8C_1e^{-8t} - 2C_2e^{-2t}$

Thus
$0 = -8C_1 - 2C_2$

From the first equation I get that
$C_2 = 1 - C_1$

Inserting this into the second equation gives:
$0 = -8C_1 - 2(1 - C_1)$

$0 = -8C_1 - 2 + 2C_1$

$0 = -6C_1 - 2$

$6C_1 = -2$

$C_1 = -\frac{1}{3}$

Thus
$C_2 = 1 - \frac{-1}{3} = \frac{4}{3}$

Thus
$x(t) = -\frac{1}{3}e^{-8t} + \frac{4}{3}e^{-2t}$

6. D'oh, algerbra mistake....its not the DE that gets ya

Thanks for slapping my math sense.

7. Originally Posted by neven87
D'oh, algerbra mistake....its not the DE that gets ya

Thanks for slapping my math sense.
No problem. I've found that most Calculus students have more problems with the algebra than the Calculus, so you're fairly typical.

-Dan

8. Well that is how it goes, i honestly didnt pay attention in high school. When you get to cal and need a refresher it is hard to go back to lower level classes ; like when i got to college and want to take pre cal and cal, my tutition wasnt going to be covered by fincial aid, so i work with what i have