# Physics related initial value problem (horizontal spring mass system)

Consider the horizontal spring-mass system where the spring-force is the only force acting on the mass. Suppose that a mass is initially at $x=x_0$ with an initial velocity $v_0$. Show that the resulting motion is the sum of two oscillations, one corresponding to the mass initially at rest at $x=x_0$ and the other corresponding to the mass initially at the equilibrium position with velocity $v_0$. What is the amplitude of the total oscillation?

I know that $x(t)=c_1\cos \omega t+c_2\sin \omega t$ and $\frac{dx}{dt}=-c_1\omega\sin \omega t+c_2\omega\cos \omega t$. I know that at $t=0$, $x(0)=x_0=c_1$ and $v_0=c_2\omega$. I also know that when $v=0$, $c_2=0$. But now I'm completely stuck on what to do from here. Any help would be greatly appreciated!

You're actually right there. So far, going by your notes, you have

$$x(t) = x_0 \cos \omega t + \frac{v_0}{\omega} \sin \omega t$$

and (incidentally) therefore

$$v(t) = -x_0 \omega \sin \omega t + v_0 \cos \omega t$$

Isn't your $x(t)$ in fact the sum of two oscillations, one represented by the cosine function, and one represented by the sine function? Check to see if one of those does indeed correspond to the mass initially at rest at $x = x_0$ (and if so, which one), and if the other corresponds to the mass initially at the equilibrium position with velocity $v_0$ (and if so, which one).

• I'm assuming by "resulting motion," the equation it's referring to is $x(t)$, right? The first part is when the mass is at rest and the second part corresponds to the mass at equilibrium with velocity $v_0$. But I don't know how to show the amplitude of the total oscillation. Sep 16, 2015 at 23:20
• You don't have to. The question only asks you to decompose the motion into two parts, each of which is one of the parts in the expression for $x(t)$ (so the answer to your question is yes, it refers to $x(t)$). Sep 16, 2015 at 23:31
• Okay. Thank you for your help! Sep 17, 2015 at 0:11