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

In summary: 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 \cos x = \frac{e^{ix} + e^{-ix}}{2}, \\\sin x = \frac{e^{ix} - e^{-ix}}{2i}.Alternatively, first multiply by dx/dt
  • #1
teme92
185
2

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.
 
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  • #2
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
[tex]\int a dt = -k/m \int x dt[/tex]
The second integral is not x^2/2.
 
  • #3
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.
 
  • #4
teme92 said:

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

Alternatively, first multiply by [itex]dx/dt[/itex] and only then integrate with respect to time. The result on taking square roots is a first-order separable ODE.
 
  • #5
Hey pasmith, thanks for the reply. So the solution you posted is equal to x(t)? The answer given is 0.5a + b
 
  • #6
One of us is confused. If [itex]a[/itex] is constant, then [tex]\frac{d^2x}{dt^2} = a[/tex] yields [tex]x(t) = \frac12 at^2 + bt + c.[/tex] However the problem which appears in the OP is [itex]ma + kx = 0[/itex], which can only be interpreted as an equation of motion if [itex]a(t) = \frac{d^2x}{dt^2}[/itex], and the solution is then [tex]x(t) = A\cos(\sqrt{\tfrac km} t) + B\sin(\sqrt{\tfrac km} t).[/tex]
 
  • #7
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.
 
  • #8
teme92 said:
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.
 

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

What is the significance of integrating ma+kx=0 to get x(t)?

Integrating ma+kx=0 allows us to find the position function, x(t), for an object undergoing simple harmonic motion. This equation represents the relationship between the mass of the object (m), its acceleration (a), and the force of the spring (kx). By integrating, we can solve for the position of the object at any given time.

How do you integrate ma+kx=0 to get x(t)?

To integrate ma+kx=0, we first rearrange the equation to isolate x. This gives us x=-(m/k)a. Then, we use the power rule of integration, which states that the integral of x^n is (x^(n+1))/(n+1). In this case, n=-1, so the integral becomes -x^(-1). Finally, we substitute -(m/k)a for x and add a constant of integration, C. This results in x(t)=-(m/k)a+C.

What are the units of the constant of integration, C?

The units of the constant of integration, C, depend on the units of the other variables in the equation. If m is measured in kilograms, k in newtons per meter, and a in meters per second squared, then the units of C will be in meters. This is because the units of x(t) must match the units of displacement, which is measured in meters.

Can this equation be used for any type of motion?

No, this equation is specifically used for simple harmonic motion, which is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium. Examples of simple harmonic motion include a mass attached to a spring and a pendulum swinging back and forth.

How does this equation relate to Newton's second law of motion?

This equation, ma+kx=0, is derived from Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration. In this equation, the force of the spring, kx, is equal to the negative of the mass times the acceleration, ma. This relationship allows us to solve for the position of an object undergoing simple harmonic motion.

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