Average rate of growth of a forced oscillator

A forced oscillator with initial displacement $x=0$ is subjected to a force $F_0 \cos\omega t$. The transient term decays to $e^{-k}$ of its original value in a time $t$. Given that the damping of the system is small, show that the average rate of growth of the oscillations is given by $$\frac{x_0}{t}=\frac{F_0}{2km\omega_0}$$where $x_0$ is the maximum steady state displacement and $\omega_0$ is the natural angular frequency of oscillation($\frac{s}{m}$)

QUESTION

I do not understand what this question is trying to say or what the solution means?

1."The transient term decays to $e^{-k}$ of its original value". So what is this value , is it the displacement or velocity?

2.if the damping is small the rate of decay becomes small and thus the solution of the rate of growth is the steady state velocity: $$\frac{-F_0}{X_m}*\sin(\omega t)$$, What am I missing, it's really frustrating!

EDIT

I am aware of the formulas given to me on the comment but i do not really understand what the average rate of growth means. What is the question really asking? Is it asking$$\frac{\int_{0}^{t} x(t) dt}{t}$$ because that won't yield the formula required?

• Have a read of this, which is concerned with x(t) ......hint as what the orginal value is hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html#c3 small damping (or underdamping) would naturally lead to a steady state solution
– user108787
Aug 12, 2016 at 18:28