Solving motion of undamped harmonic oscillator

[jlmac2001]

The question is: Solve for the motion of the undamped harmonic oscillator with an applied force F, treated in class, when the force is no longer constant but has the form F=F0+kT, where Fo and k are constants. Use the intial conditions x(0)=d and x'(0) =v0.

I'm trying to solve this problem using my notes but I don't see anything like it. What I'm thinking is that I need to find an equation of motion. Something like x(t)=x''+w0^2x=0, x(0)=x''+w0^2x=d, x'(0)=x''+w0^2x=v0. Help me!

 

[cookiemonster]

You will need to end up with an equation of motion, but you can't start with one. You got to start with what you know, and all you know is force.

Use

[tex]F = m\ddot{x}[/tex]

where F is your given applied force. This gives you a differential equation that you must solve.

cookiemonster

 

[M98Ranger]

To solve this problem, we can use the equation of motion for an undamped harmonic oscillator, which is:

x'' + ω0^2x = F/m

where x is the displacement, ω0 is the natural frequency, and F is the applied force. In this case, the force is no longer constant but has the form F = F0 + kT, where F0 and k are constants.

To find the solution, we first need to find an expression for F/m. Since we know that F = F0 + kT, we can substitute this into the equation of motion:

x'' + ω0^2x = (F0 + kT)/m

Next, we can rearrange this equation to isolate x'' on one side:

x'' = -(ω0^2/m)x + (F0/m) + (kT/m)

Now, we can use the initial conditions given in the problem to find the specific values for x(0) and x'(0). Substituting x(0) = d and x'(0) = v0 into the equation above, we get:

d'' = -(ω0^2/m)d + (F0/m) + (kT/m)
v0'' = -(ω0^2/m)v0 + (F0/m) + (kT/m)

We can solve these equations for d'' and v0'' and substitute them back into the equation of motion:

x'' = -(ω0^2/m)x + (F0/m) + (kT/m)
= -(ω0^2/m)x + d'' + (ω0^2/m)d + (kT/m)
= -(ω0^2/m)(x-d) + (kT/m)

Now, we can solve for x(t) by integrating the equation twice:

x(t) = -(ω0^2/m)∫(x-d)dt + (kT/m)t + c1
= -(ω0^2/m)(x-d)t + (kT/m)t^2/2 + c1t + c2

Finally, we can use the initial conditions to solve for the constants c1 and c2:

x(0) = d = c2
x'(0) = v0 = -ω0^2c2 + kT/m
c1 = (v0 + ω0^2d