# Root-mean-square of a sine wave

by xlq
Tags: rootmeansquare, sine, wave
 P: 2 rms of sine wave = peak * 1/SQRT(2) how is this derived from the rms equation? (Search engines have returned no useful results.)
 P: 406 The original sine wave is take to represent a current. This current is squared to get the power. P=I^2. Integrating sine squared between from time 0 to pi gives; $$\int_0^{\pi} \sin^2{t} dt = \frac{\pi}{2}$$ This is the same a multiplying the constant power 1/2 by the time pi seconds. So 1/2 is the "average" power level, and since I = sqrt(P), 1/sqrt(2) is the "average" current level needed to obtain a meaningful powerlevel result.
 P: 2 OK thanks. One more question: how did you integrate the sine-squared?
P: 227

## Root-mean-square of a sine wave

use the trig identity sin^2(t)=1/2(1-cos(2t) (it might be +, i can't remember. Derive it from the double angle formula for cos(2t)) :)
Sci Advisor
P: 5,776
 OK thanks. One more question: how did you integrate the sine-squared?
sin^2 + cos^2 =1. The rest is trivial.
 PF Patron HW Helper Sci Advisor P: 4,755 how do u use sin^2 + cos^2 =1 to evaluate the integral of sinē mathman?
P: 406
 Quote by mathman sin^2 + cos^2 =1. The rest is trivial.
No. That was the trivial part. The rest is hard.

Using the double angle cossine formula;

$$\cos{2\theta} = cos^2{\theta} - sin^2{\theta} = 1 - 2\sin^2{\theta}$$
Therefore;
$$2\sin^2{\theta}= 1-\cos{2\theta}$$
$$\Rightarrow \sin^2{\theta}= \frac{1}{2} \left(1-\cos{2\theta} \right)$$

So that means the integral can be evaluated as follows.

$$\int_0^{\pi} \sin^2{\theta} d\theta = \int_0^{\pi} \frac{1}{2} \left(1-\cos{2\theta} \right) d\theta$$
$$= \frac{1}{2} \int_0^{\pi}1 d\theta -\frac{1}{2} \int_0^{\pi} \cos{2\theta} d\theta$$
$$= \frac{1}{2} \left[ \theta |_0^{\pi}\right] + \frac{1}{4}\left[ \sin{2\theta}|_0^{\pi}\right]$$
$$= \frac{1}{2} ( \pi - 0) \frac{1}{4} ( 0 - 0) = \frac{\pi}{2}$$
 HW Helper P: 2,566 sin2(x) is just cos2(x) shifted by pi/2, and each have a period of pi, so their integrals over 0 to pi are equal. Thus: $$\pi=\int_0^\pi dx = \int_0^\pi (\sin^2 x + \cos^2 x) dx = 2 \int_0^\pi \sin^2 x dx$$
Sci Advisor
P: 5,776
 how do u use sin^2 + cos^2 =1 to evaluate the integral of sinē mathman?
Integral of sin^2 = integral of cos^2, so each must be 1/2 of the integral of 1.
 P: 1 Schaums - Theory and Problems of Electric Circuits (4nd ed) 1.3- Solution $$\copyright$$ Sin squared = 1 - cosine 2x 1.3 A certain circuit element has a current i = 2.5 sin $$\omega$$t (mA), where $$\omega$$ is the angular frequency in rad/s, and a voltage difference v = 45 sin $$\omega$$t (V) between terminals. Find the average power $$P_{avg}$$ and the energy $$W_{T}$$ transferred in one period of the sine function. RECALL THAT : VOLTAGE(v) X CURRENT(i) = WORK($$W_{T}$$) 2.5 X 45 = 112.5 so.... $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} vi dt = W_{T} = 112.5 \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt$$ RECALL TRIGONMETRY IDENTITY $$sin^2x = 1/2(1- cos2x)$$ $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt = W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ WE NOW HAVE .... $$W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ CALCULUS IDENTITY FOR $$\int\cos(x) dx = \sin x + C$$ PROOF with U Substitution : Let u = 2x THEN $$\frac{du}{dx} = 2$$ AND dx = $$\frac{1}{2}du$$ $$\int\cos\2x$$ = $$\int\cos u\frac{1}{2}du$$ $$\frac{1}{2}\int\cos u du$$ $$\frac{1}{2}\sin u + C$$ $$\frac{1}{2}\sin2x + C$$ SO NOW .... $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{2\pi}{\omega}}$$ NOW UPPER AND LOWER INTEGRATION GIVES $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{}}$$ SUBSTITUTION HERE .... AFTER SUBSTITUTION RECALL TRIGONOMETRY SINE $$\frac{\pi}{\theta}$$ = 0 $$\left[ \frac{1}{2} ( ( \frac{2\pi}{\omega} ) - \frac{1}{2} \sin 2 ( \frac{2\pi}{\omega} ) ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( ( 0 ) - \frac{1}{2} \sin 2 ( 0 ) ) \right]_{0}^{\frac{}}$$ OK ... THIS IS WHAT'S LEFT $$W_{T} = 112.5 \left[ ( \frac{1}{2} ) ( \frac{2\pi}{\omega} ) - 0 \right]$$ THEREFORE: $$W_{T} = 112.5 \left[ ( \frac{\pi}{\omega} ) \right] = ( \frac{ 112.5\pi}{\omega} ) \right]$$ FINALLY: $$P_{avg} = ( \frac{ W_{T} }{ \frac { 2\pi}{\omega}} ) = 56.25mW$$
 P: 2 The answers given above by ObsessiveMathsFreak, namely that the average is $$\frac{\pi}{2}$$ is wrong. The integral over a period will *not* give the average, merely the area under the curve, which should be obvious to everyone who provided the answers - as proof consider that the average of $$\sin^2(\theta)$$ over any whole period should be the same, be it from 0 to $$\pi$$ or from 0 to $$500\pi$$. Yet, with the above answers this is clearly not the case. A (or The) correct way to find the average is via average value theorem. $$f_{avg}(x)=\frac{1}{b-a}\int_{a}^{b}f(x)$$ Which in the case of $$\sin^2(\theta)$$ yields $$f_{avg}=\frac{1}{2}$$
 P: 2 I have a problem that has been bugging me all year. Yes, all year - which i know is sad. can anyone tell me how to solve the following problem? What is the power applied to a string that is 10 meters long, has a hertz of 440? I can't really find a formula that i can understand. i don't know much about physics and this was a problem from last semesters final, as extra credit on a algebra test - go figure. anyway it is still bugging me. thanks,
 P: 2 Are you sure that is the whole question? That can't possibly be enough information to come up with an answer, surely... Without knowing anything about the string, we could assume that it was a ideal (lossless) carrier of the wave, so the "power" would be the energy taken to excite it - however we aren't given the amplitude of the wave. I'm assuming that the system is a string attached to a fixed point, being excited by the other end? The question seems far to vague to have a numeric answer, and if they just wanted proof of understanding why did they give exact length/frequency instead of length of L, frequency of F Hz or such...
 P: 2 i think i might have left off a part...something about the wavelength being 2 meters...woudl that help? if not then i will just have to try to get the question again. thanks for looking at this for me ;-)

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