- #1
kafn8
Homework Statement
A particle of mass m is initially at rest at x = 0.
It is acted upon by a force F = A cosh (\beta t) (1)
A) Show that at very small values of t, the position is approximately given by x(t) = \frac{1}{2}\frac{F_0}{m}t^2 (2),
where F_0 is the force at t =0
Homework Equations
The Attempt at a Solution
If F(t) = Acosh(\beta t) = ma(t)
then a(t) = \frac{A}{m}cosh(\beta t)
Integrating twice yields the position such that
x(t) = \frac{A}{\beta^2 m}\left[ cosh(\beta t) - 1 \right], (3)
Also, F_0(t=0)=m\left[\frac{A}{m}cosh(\beta(0)) \right]=A
With that out of the way, I've tried taking the limit of (3) as t \rightarrow 0 but end up with the following:
$$\lim_{t \rightarrow 0}x(t) = \lim_{t \rightarrow 0} \frac{A}{\beta^2 m}\left[ cosh(\beta (0)) - 1 \right]$$
$$= \frac{A}{\beta^2 m}\left[ (1) - 1 \right]$$
$$= 0$$
But all this says is that for very small values of time, the particle barely moves away from x=0. This does not directly confirm that (2) is a good approximation. Any guidance is much needed and greatly appreciated!