# Simple Harmonic Motion Question

1. May 24, 2012

### sketchd2

1. The problem statement, all variables and given/known data
A particle moves with SHM in a straight line.
In the first second after starting from rest,it travels a distance of 1m in a constant direction.
In the next second, it travels a distance of 2m in the same direction.
Find its maximum acceleration

2. Relevant equations
x=acos(ωt)
$\dot{x}$=-ω√(a2-x2)
$\ddot{x}$=-ω2x

3. The attempt at a solution
Starting from rest ∴ when $\dot{x}$=0, x=a, $\ddot{x}$=maximum.
maximum $\ddot{x}$=-ω2x
maximum amplitude = 1.5
∴ amplitude $\geq$1.5

After many more lines of work, no solution was able to be found.
Can someone help?

2. May 24, 2012

### genericusrnme

You have two equations and two unknowns, you don't need to do anything more than plugging values into x=acos(ωt).

From the given information what values do you suppose you need to plug in?

3. May 24, 2012

### sketchd2

when t=1, x=a-1
when t=2, x=a-3

4. May 24, 2012

### genericusrnme

In the first second after starting from rest,it travels a distance of 1m in a constant direction.
In the next second, it travels a distance of 2m in the same direction.

I'd use this, what exactly is this saying?

5. May 24, 2012

### sketchd2

it is saying that after the 1st second the particle has moved 1m from the amplitude position and that after the 2nd second the particle mas moved a total of 3m from the amplitude position.

6. May 24, 2012

### vela

Staff Emeritus
You have
\begin{align*}
a - a \cos \omega &= 1 \\
a - a \cos 2\omega &= 3
\end{align*} Solve for a in the first equation and substitute into the other. Then use a trig identity for $\cos 2\omega$ to write it in terms of $\cos \omega$.

7. May 25, 2012

### sketchd2

ok. thank you for this. it helps me a lot