Question #255963

For the function v = 12 sin 40, calculate the: a) mean b) root mean square (RMS) 

Over a range of 0 ≤ Ø ≤ π/4 radians. 


(Note: the trigonometric identity cos 2Ø = 1-2sin^2 Ø)


1
Expert's answer
2021-10-25T16:32:59-0400
vav=124π0π4sin4θdθ=34π0πsinxdxv_{av}=12\frac{4}{\pi}\int _0^{\frac{\pi}{4}} \sin{4\theta}d\theta=3\frac{4}{\pi}\int _0^{\pi} \sin{x}dxvav=3(cosπcos0)4π=24πv_{av}=-3(\cos{\pi}-\cos{0})\frac{4}{\pi}=\frac{24}{\pi}




vrms2=1224π0π4sin24θdθ=1222π0π4(1cos8θ)dθv_{rms}^2=12^2\frac{4}{\pi}\int _0^{\frac{\pi}{4}} \sin^2{4\theta}d\theta=12^2\frac{2}{\pi}\int _0^{\frac{\pi}{4}} (1-\cos{8\theta})d\theta0π4(1cos8θ)dθ=π41802πsinydy=π4\int _0^{\frac{\pi}{4}} (1-\cos{8\theta})d\theta=\frac{\pi}{4}-\frac{1}{8}\int _0^{2\pi} \sin{y}dy=\frac{\pi}{4}

Thus,



vrms2=1224π0π4sin24θdθ=1222ππ4v_{rms}^2=12^2\frac{4}{\pi}\int _0^{\frac{\pi}{4}} \sin^2{4\theta}d\theta=12^2\frac{2}{\pi}\frac{\pi}{4}vrms=122v_{rms}=\frac{12}{\sqrt{2}}

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