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?
Follow Math Help Forum on Facebook and Google+
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
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.
What is ? I dunno...
Originally Posted by fardeen_gen What is ? I dunno... $\displaystyle \cos(\omega{t})^2=\frac{1+\cos(2\omega{t})}{2}$ So we have $\displaystyle \frac{1}{2}\int\bigg[1+\cos(2\omega{t})\bigg]dt=\frac{1}{2}\bigg[t+\frac{1}{2\omega}\sin(2\omega{t})\bigg]+\bold{\color{red}C}$
Originally Posted by Mathstud28 ...$\displaystyle +\bold{\color{red}C}$ Aha! you remembered!
Originally Posted by Jhevon Aha! you remembered! I can't believe you got that
View Tag Cloud