# Homework Help: Simple harmonic motion - initial acceleration

1. Nov 16, 2015

### Tzabcan

1. The problem statement, all variables and given/known data
To work out the intial acceleration, do we just use the equation:

So at t = 0

We eliminate the wt inside the bracket, and are left with Aω^(2)sin(Φ + π)

2. Relevant equations

3. The attempt at a solution

The part which I'm not so sure on, is if i have values for Φ, do i literally treat π as it's true value or 180?

Say Φ = π/2 (just a random number)...and A= 9 and ω = 3

Will this give me the initial acceleration:

((9* 3^(2)) sin (π/2+ 180) = ((9*3^(2))sin(270)

?

But with my actual example i have, doing it this way gives me a huge massive negative number as the sin of 270 is -1 lol.

Thanks

Last edited: Nov 16, 2015
2. Nov 16, 2015

### SteamKing

Staff Emeritus
So, you're saying that -81 is a 'huge massive negative number'? Does A have any units attached?

3. Nov 16, 2015

### Mister T

What does it mean to "work out the initial acceleration"? To understand that question we need to know what is given. That's the reason for the template.

4. Nov 17, 2015

### J Hann

It looks like pi (radians) was inserted in this equation to account for the negative sign.
For SHM if we omit the phase angle phi
x(t) = A sin w t
v(t) = A w cos w t
a(t) = - A w^2 sin w t = A w^2 sin (w t + pi)
Since sin (theta + pi) = - sin theta

5. Nov 25, 2015

### Tzabcan

Well, what we have been given is ω = 2.5 x 10^3 s A = 1.8mm Φ = π/2

I'm just confused how I'm supposed to be using this equation.

Do i quite literally just do:

0.0018m * (2.5 x 10^3)^2 sin (π/2 +π) ?

And then take the inside to be sin (270)? or do i take it as 4.7? :s

Thanks

6. Nov 25, 2015

### Mister T

sin(270°) = -1

7. Nov 25, 2015

### Tzabcan

Oh haha, should've put it into the calculator and see for myself. Thanks :)

8. Nov 25, 2015

### Mister T

Or, you could keep track of the units as you go, something you seem to not be doing. It will catch up with you!