Undamped oscillator with driving force

[thenewbosco]

just wondering how to go about this one:
An undamped harmonic oscillator is subject to a driving force $$F_0e^{-bt}$$. It starts from rest at the origin (x=0) at time t=0.

assuming a general solution $$x(t)=A cos(\omega t - \phi) + Be^{-\alpha t}$$ where A, $$\omega$$, $$\phi$$, B, and $$\alpha$$ are real constants, find the position x(t) as a function of time.

I was thinking to take this general solution, differentiate, and apply initial conditions to the two equations, however all i could solve for is alpha in terms of omega.
Then i thought to differentiate again and plug into the equation
$$ma=-kx + F_{driving}$$ however i cannot see how i can solve for all the constants with these two equations..any help?

 

[Nate810]

To solve this problem, you need to first solve the differential equation that describes the motion of the oscillator. This is done by taking the second derivative of the position with respect to time, which yields the equation:m\ddot{x} + kx = F_0e^{-bt}where m is the mass of the oscillator, k is the spring constant, and b is the damping coefficient.Assuming a solution of the form x(t) = A\cos(\omega t - \phi) + Be^{-\alpha t}, where A, \omega, \phi, B, and \alpha are real constants, we can substitute it into the differential equation and solve for the constants. After some algebraic manipulation, we can find that:A = \frac{F_0}{\sqrt{k^2 + (m\omega)^2}}\omega = \sqrt{\frac{k}{m}}\phi = \arctan\left(\frac{m\omega}{k}\right)B = 0\alpha = bTherefore, the solution of the position of the undamped harmonic oscillator under this driving force is:x(t) = \frac{F_0}{\sqrt{k^2 + (m\omega)^2}}\cos(\sqrt{\frac{k}{m}}t - \arctan\left(\frac{m\omega}{k}\right))

 

[kyle8921]

I would suggest approaching this problem by using the equations of motion for an undamped harmonic oscillator with a driving force. This would involve using the second-order differential equation:

m\ddot{x} + kx = F_0e^{-bt}

Where m is the mass of the oscillator, k is the spring constant, and F_0e^{-bt} is the driving force. From this equation, we can find the general solution for x(t) as:

x(t) = A\cos(\omega t - \phi) + \frac{F_0}{k-b^2m}e^{-bt}

Where A and \phi are determined by the initial conditions of the oscillator. From this solution, we can see that the position of the oscillator is a combination of the natural oscillation (A\cos(\omega t - \phi)) and the forced oscillation (\frac{F_0}{k-b^2m}e^{-bt}). The amplitude and phase of the natural oscillation are determined by the initial conditions, while the amplitude of the forced oscillation is determined by the driving force and the parameters of the oscillator (mass, spring constant, and damping coefficient).

To solve for the constants A and \phi, we can use the initial conditions given in the problem (starting from rest at the origin at t=0). This would give us the following equations:

x(0) = A + \frac{F_0}{k-b^2m} = 0

\dot{x}(0) = -\omega A + b\frac{F_0}{k-b^2m} = 0

Solving these equations for A and \phi, we get:

A = \frac{-F_0}{k-b^2m}

\phi = \tan^{-1}\left(\frac{b\omega}{k-b^2m}\right)

Substituting these values back into the general solution for x(t), we get:

x(t) = \frac{-F_0}{k-b^2m}\cos(\omega t - \tan^{-1}\left(\frac{b\omega}{k-b^2m}\right)) + \frac{F_0}{k-b^2m}e^{-bt}

Therefore, the position x(t) as a function of time can be determined using this equation. It is important to note that the constants A, \