- #1

kafn8

## Homework Statement

A particle of mass m is initially at rest at x = 0.

It is acted upon by a force [itex]F = A cosh (\beta t)[/itex] (1)

A) Show that at very small values of t, the position is approximately given by [itex]x(t) = \frac{1}{2}\frac{F_0}{m}t^2[/itex] (2),

where [itex]F_0[/itex] is the force at [itex]t =0[/itex]

## Homework Equations

## The Attempt at a Solution

If [itex]F(t) = Acosh(\beta t) = ma(t)[/itex]

then [itex]a(t) = \frac{A}{m}cosh(\beta t)[/itex]

Integrating twice yields the position such that

[itex]x(t) = \frac{A}{\beta^2 m}\left[ cosh(\beta t) - 1 \right][/itex], (3)

Also, [itex]F_0(t=0)=m\left[\frac{A}{m}cosh(\beta(0)) \right]=A[/itex]

With that out of the way, I've tried taking the limit of (3) as [itex]t \rightarrow 0[/itex] 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!