Simple harmonic motion - initial acceleration

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Tzabcan
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Homework Statement


To work out the intial acceleration, do we just use the equation:

Screenshot_2.jpg


So at t = 0

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

Homework Equations

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
 
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Tzabcan said:

Homework Statement


To work out the intial acceleration, do we just use the equation:

Screenshot_2.jpg


So at t = 0

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

Homework Equations

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

Homework Statement


To work out the intial acceleration, do we just use the equation:

Screenshot_2.jpg


So at t = 0

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

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

270° ≈ 4.7 radians.

sin(270°) = -1

sin(4.7 rad) ≈ -1

So, to answer your question, yes. You can use either.
 
Mister T said:
270° ≈ 4.7 radians.

sin(270°) = -1

sin(4.7 rad) ≈ -1

So, to answer your question, yes. You can use either.

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