I've started this, but I'm getting a bit stuck.
When a mass of $2kg$ is attached to a spring whose spring constant is $32 N/m$ it comes to a rest in the equilibrium position. Starting at $t=0s$, a force equal to $Q(t)=68e^{-2t}cos(4t)$ is applied to the system. Find the equation of motion in the absence of damping.
$$m\ddot x +\beta \dot x + kx=Q(t), 2\lambda=\frac{\beta}{m}=0,\omega_0^2=\frac{k}{m}=16$$
$$\ddot x +16x=68e^{-2t}\cos(4t)$$
$$x_c=c_1\cos(\omega_0 t)+c_2\sin(\omega_0 t)=c_1\cos(4t)+c_2\sin(4t)$$

Question

I've started this, but I'm getting a bit stuck.
When a mass of $2kg$ is attached to a spring whose spring constant is $32 N/m$ it comes to a rest in the equilibrium position. Starting at $t=0s$, a force equal to $Q(t)=68e^{-2t}cos(4t)$ is applied to the system. Find the equation of motion in the absence of damping.
$$m\ddot x +\beta \dot x + kx=Q(t), 2\lambda=\frac{\beta}{m}=0,\omega_0^2=\frac{k}{m}=16$$
$$\ddot x +16x=68e^{-2t}\cos(4t)$$
$$x_c=c_1\cos(\omega_0 t)+c_2\sin(\omega_0 t)=c_1\cos(4t)+c_2\sin(4t)$$

anonymous · Answer

Correction on one of the equations I noticed now.
$$\ddot x + 16x = 34e^{-2t}\cos(4t)$$
With this I was able to work out $x_p$
$$x_p=\frac{1}{2}e^{-2t}\cos(4t) -2e^{-2t}\sin(4t)$$
This gives:
$$x=c_1\cos(4t)+c_2\sin(4t) + \frac{1}{2}e^{-2t}\cos(4t)-2e^{-2t}\sin(4t)$$
Not sure how to find the constants from the problem description though.

sidsiddhartha · Answer

u have two constants here so u have to differentiate $ x$ two times

anonymous · Answer

Okay, so I see the first I think $Q(0)=m\ddot x$ I also see that as time goes on ($\lim_{t \to \infty}$)? $x\rightarrow 0$. Not sure I know how to take care of that one, either that or I don't remember how to. If so, $x_p$ goes toward 0, so the components of $x_c$ would need to be opposite of each other.