Forced Oscillations: +ve Sign of Fcoswt

[dennisrobert]

In the equation of forced oscillations (given below),

ma = -kx + FCoswt, why is that the 2nd term on the right hand side is given the +ve sign. I know that kx should be negative since 'ma' and 'x' are in opposite directions. I can't quite seem to get the gist of Fcoswt being given a +ve sign. Can someone clarify?

 

[Isak BM]

It's arbitrary. Replacing the $+$ with a $-$ would work just as well. It would only change the phase of the forced oscillations. It would effectively change the direction in which the mass is being displaced instantly after $t = 0$. If the sign is $+$, then the mass gets displaced (originally, directly after $t = 0$) along the positive x-axis, and if the sign is $-$ then the original displacement is along negative x-axis.

 

[vanhees71]

The second term is simply the time dependent external force. It can have any sign and also any functional shape as you like. The harmonic shape is particularly interesting since you can describe any external force by a Fourier series (periodic force) or a Fourier integral (non-periodic force).

The general solution for any external force is given with help of the retarded Green's function of the corresponding differential operator of the homogeneous equation, i.e., which obeys the equation of motion with a $\delta$ force,

$$(\partial_t^2+\omega_0^2) G(t,t')=\delta(t-t'). \qquad (1)$$

If you can find the Green's function, for any external force, a solution reads

$$x(t)=\int_{-\infty}^{\infty} \mathrm{d} t' G(t,t') \frac{F(t')}{m}.$$

To find the Green's function, you make the ansatz

$$G(t,t')=\begin{cases} 0 & \text{for} \quad t<t', \\<br /> g(t,t') & \text{for} \quad t>t'.<br /> \end{cases}$$

The imposed condition for the Green's function to vanish for $t<t'$ ensures that the external force only acts from the past and not the future on the position of the particle as it must be according to causality. That's what distinguishes the retarded propagator from any other solution of the Green's-function equation.

To solve it, you can simply use the fact that for $t\neq t'$ the $\delta$ distribution vanishes, and then the solution reads

$$g(t,t')=A \exp(\mathrm{i} \omega_0 t)+B \exp(-\mathrm{i} \omega_0 t). \qquad (2)$$

Now we need two initial conditions to determine the integration constants, A and B. The first is given by the continuity of $G(t,t')$ at $t=t'$, i.e., we must have

$$g(t'+0^+,t')=0. \qquad (3)$$

The second condition we find by integrating (1) over an infinitesimal interval $t \in (t'-0^+,t'+0^+)$, leading to the condition on the jump of the first derivative:

$\partial_t G(t'+0^+,t')-\partial_t G(t'-0^+,t')=\partial_t G(t'+0^+,t')=\partial_t g(t'+0+,t') \stackrel{!}{=} 1. \qquad (4)$

Plugging (3) and (4) in (2) leads to

$$A+B=0, \quad A-B=\frac{1}{\mathrm{i} \omega_0} \; \Rightarrow A=-B=\frac{1}{2 \mathrm{i} \omega_0}.$$

This finally gives

$$G(t,t')=\Theta(t-t') \frac{\sin(\omega_0 t)}{\omega_0}.$$

 

[AlephZero]

The equation is just "force = mass x acceleration", and positive force and positive acceleration are in the same direction.

When you solve "elementary" questions in dynamics by you can often draw a diagram which measures individual forces, displacements, accelerations, etc as positive in any directions you like, and then figure out the correct equations by looking at the diagram

In more advanced work it is much easier to use the same coordinate system for everything.
That is why the force from the spring is -kx measured in the same coordinate system as x, in this equation.