Trajectory of a free particle in a sinusoidal force field

Question

I'm struggling to derive the analytical form for the initial (transient) motion of a free particle in a sinusoidal force field.

For the sake of simplification, I'm operating in a one-dimensional space. A particle of mass $\mathbf{m}$ is initially at $\mathbf{x}=0$ and at rest $(\mathbf{v}=0)$. At $\mathbf{t}=0$, the particle begins to be subject to an oscillating force field: $$f(t) = a\sin(\omega t)$$

If I proceed by integrating the force, taking into account the initial zero velocity I get: $$v(t)=\frac{a}{m\omega}(1-\cos(\omega t))$$ and thus for the position: $$x(t)=\frac{a}{m\omega}(t-\frac{sin(\omega t)}{\omega})$$ What I find un-intuitive about my solution is that the velocity is always positive, with as a consequence the particle moving further and further away from the origin. Given that within each period the total positive acceleration equals the total negative acceleration, I would have expected the particle to stabilize in a "periodic" trajectory.

Am I making a blatant mistake in my integration, or is my intuition wrong - and, if it's wrong, what is the physical interpretation of the velocity being always positive even though the particle is subject to an acceleration which always nets to zero?

Answer 1

I think that your result is correct. Concerning the interpretation, I would say the following. The average acceleration over a time $T = 2\pi/\omega$ is zero, the average velocity is $a/(m\omega)$, and the average position is $at/(m\omega)$. These averages can been obtained by applying $(1/T) \int_0^T dt ...$

So you see that on average everything is consistent: the acceleration vanishes, the velocity is constant and the position grows linearly with time: the motion is straight and uniform. In other words, having zero acceleration doesn't imply having zero velocity, right?

Furthermore you can also check that the average energy transfered into the system by the force $f(t)$ is zero (simply compute $(1/T) \int_0^T dt f(t) v(t) = 0$)

Answer 2

The problem is in your initial conditions. Your second order differential equation reads, $$ m\ddot{x}(t)=\sin(\omega t) $$ Consider the homogeneous solution of $$ \ddot{x}(t)=0\\ \dot{x}(t)=C\\ x(t)=Ct+D $$ From the initial conditions both $C$ and $D$are zero. The solution is the sum of the homogeneous solution and the particular solution. The particular solution of $m\ddot{x}(t)=\sin(\omega t)$ is $$ \frac{1}{m}(A\cos(\omega t) + B\sin(\omega t)) $$ Using the initial conditions to evaluate $A$ and $B$ $$ x(0)=-\frac{A}{m}=0\\ A=0\\ \dot{x}(0)=\frac{\omega B}{m}=0\\ B=0 $$ Since the particle had no momentum at $t=0$ and you didn't apply a force to the particle at $t=0, x=0$ it will just sit still forever.