# Thread: General solution of this equation

1. ## General solution of this equation

I need to get a general solution of the following:

ma +5kx = -mgsin(alpha) +11klo

m = mass, k stiffness, g gravity, lo natural length of spring
I think that I need it in the form of:

ab^2 + ab + c = 0

so i can get the auxiliary equation and therefore find the general solution but Im not sure how to rearrange to get in this form

cheers!

2. Originally Posted by thermalwarrior
I need to get a general solution of the following:

ma +5kx = -mgsin(alpha) +11klo

m = mass, k stiffness, g gravity, lo natural length of spring
I think that I need it in the form of:

ab^2 + ab + c = 0

so i can get the auxiliary equation and therefore find the general solution but Im not sure how to rearrange to get in this form

cheers!
Well,
$\displaystyle a = \frac{d^2x}{dt^2}$, so

$\displaystyle m \cdot \frac{d^2x}{dt^2} + 5kx = -mg~sin(\alpha) + 11kl_0$
if that's what you are looking for.

-Dan

[Pet Peeve]
g is "the acceleration due to gravity." Gravity is a force, not an acceleration.
[/Pet Peeve]

3. Right I understand that bit but its the next part Im stuck on!

4. Originally Posted by topsquark
$\displaystyle m \cdot \frac{d^2x}{dt^2} + 5kx = -mg~sin(\alpha) + 11kl_0$
Originally Posted by thermalwarrior
Right I understand that bit but its the next part Im stuck on!
The auxiliary equation for the homogeneous solution is
$\displaystyle mq^2 + 5k = 0$

So
$\displaystyle q = i \sqrt{\frac{5k}{m}}$

which makes the general solution to the homogeneous equation
$\displaystyle x(t) = A~sin \left ( t \sqrt{\frac{5k}{m}} \right ) + B~cos \left ( t \sqrt{\frac{5k}{m}} \right )$

Now find a particular solution and apply your initial conditions.

-Dan

5. i think i understand now. but where does the t variable come into it?

6. Originally Posted by thermalwarrior
i think i understand now. but where does the t variable come into it?
The general solution of your auxiliary equation is $\displaystyle q_1, q_2$, so the general solution to your homogeneous equation is
$\displaystyle x(t) = Ae^{q_1t} + Be^{q_2t}$

I'm not sure exactly what you are confused about? t is simply your independent variable in terms of which you are solving for x.

-Dan