Integrating ma+kx=0 to get x(t)

[teme92]

Homework Statement

As stated in the title, I'm having trouble integrating ma+kx=0 to get x(t)

Homework Equations

The Attempt at a Solution

So I know I have to integrate twice but I'm not getting the answer required.

∫a = -k/m∫x

v = (-k/m)[(x²/2) + C]
∫v = (-k/2m)∫x² + (-kC/m)
x = (-k/2m)x³/3 + (-kC/m)

This is clearly wrong because I have no time variables. Any help would be much appreciated.

 

[nasu]

Integral of a is meaningless if you don't specify the variable of integration. Same for integral of x.
And you need to integrate over the same variable on both sides to maintain the equality.
If you integrate over time you have
\int a dt = -k/m \int x dt
The second integral is not x^2/2.

 

[teme92]

Hey nasu thanks for the reply.

∫xdt = xt +C

Then after integrating the second time:

∫vdt = (-k/m)∫xtdt
x = (-k/m)xt²/2 +C

I'm unsure about the constant of integration and whether there should be two or not.

 

[pasmith]

Homework Statement

As stated in the title, I'm having trouble integrating ma+kx=0 to get x(t)

Homework Equations

The Attempt at a Solution

So I know I have to integrate twice but I'm not getting the answer required.

∫a = -k/m∫x

v = (-k/m)[(x²/2) + C]
∫v = (-k/2m)∫x² + (-kC/m)
x = (-k/2m)x³/3 + (-kC/m)

This is clearly wrong because I have no time variables. Any help would be much appreciated.

You are trying to solve \frac{d^2 x}{dt^2} + \frac km x = 0. This is a second-order linear homogenous ODE with constant coefficients, so its solution is Ae^{\lambda_1 x} + Be^{\lambda_2 x} where \lambda_1, \lambda_2 are the roots of \lambda^2 + \frac km = 0. You will want to use the relations <br /> \cos x = \frac{e^{ix} + e^{-ix}}{2}, \\<br /> \sin x = \frac{e^{ix} - e^{-ix}}{2i}.<br />

Alternatively, first multiply by dx/dt and only then integrate with respect to time. The result on taking square roots is a first-order separable ODE.

 

[teme92]

Hey pasmith, thanks for the reply. So the solution you posted is equal to x(t)? The answer given is 0.5a + b

 

[pasmith]

One of us is confused. If a is constant, then \frac{d^2x}{dt^2} = a yields x(t) = \frac12 at^2 + bt + c. However the problem which appears in the OP is ma + kx = 0, which can only be interpreted as an equation of motion if a(t) = \frac{d^2x}{dt^2}, and the solution is then x(t) = A\cos(\sqrt{\tfrac km} t) + B\sin(\sqrt{\tfrac km} t).

 

[teme92]

Well I remember from class that the constant of integration can basically be removed from the equation depending on where you set your reference frame. I'll ask my professor to clarify the question tomorrow in class. Thanks for the help guys.

 

[nasu]

Hey nasu thanks for the reply.

∫xdt = xt +C

Then after integrating the second time:

∫vdt = (-k/m)∫xtdt
x = (-k/m)xt²/2 +C

I'm unsure about the constant of integration and whether there should be two or not.

No, x is not a constant but a function of t. You cannot integrate that way.
This is why you have to do what pasmith showed you.
And the constant of integration cannot avoid this.