1. ## [SOLVED] Integration?

If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

A)1/2 m*ω^2*A^2
B) zero
C) 1/4 mω^2 A^2
D) 1

I got B) zero as the answer.
Is it correct?

2. Originally Posted by fardeen_gen
If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

A)1/2 m*ω^2*A^2
B) zero
C) 1/4 mω^2 A^2
D) 1

I got B) zero as the answer.
Is it correct?
i am pretty sure B) is wrong. running it through in my head (not sure how confident you want to be in that ) i got C)

do you get zero when you do $\displaystyle \int_0^T \cos^2 \omega t~dt$ ? nope. so the answer can't be zero

3. Originally Posted by fardeen_gen
If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

A)1/2 m*ω^2*A^2
B) zero
C) 1/4 mω^2 A^2
D) 1

I got B) zero as the answer.
Is it correct?
I get c, I am not sure how you would get zero for this answer.

4. What is ? I dunno...

5. Originally Posted by fardeen_gen
What is ? I dunno...
$\displaystyle \cos(\omega{t})^2=\frac{1+\cos(2\omega{t})}{2}$

So we have

$\displaystyle \frac{1}{2}\int\bigg[1+\cos(2\omega{t})\bigg]dt=\frac{1}{2}\bigg[t+\frac{1}{2\omega}\sin(2\omega{t})\bigg]+\bold{\color{red}C}$

6. Originally Posted by Mathstud28
...$\displaystyle +\bold{\color{red}C}$
Aha! you remembered!

7. Originally Posted by Jhevon
Aha! you remembered!
I can't believe you got that