12
$\begingroup$

I have problem with calculating average of sine. In my book, there is claim, that $$\langle\sin^2\omega t\rangle=\frac12$$ Now I have a question how to get this result mathematically? I know you must integrate by cycle, but I can not get to this result

$\endgroup$

migrated from physics.stackexchange.com May 21 '14 at 21:35

This question came from our site for active researchers, academics and students of physics.

  • 1
    $\begingroup$ Hint: there is an identity for writing $\sin^2{x}$ in terms of a constant plus $\cos(2x)$. $\endgroup$ – The Photon May 21 '14 at 20:56
32
$\begingroup$

There is a nice trick. You know that:

$$\sin^2 \omega t+\cos^2 \omega t = 1$$

Calculate the average of this equalty, since the average over a cycle is the same for the sine and the cosine and $\langle 1 \rangle = 1$:

$$\langle \sin^2 \omega t \rangle= \frac 1 2$$

$\endgroup$
8
$\begingroup$

\begin{align} &\frac{1}{T}\int_{-T/2}^{+T/2}\sin^2\omega t\,dt\\ &\overset{1-2\sin^2\omega t=\cos2\omega t}{=} \frac{1}{T}\int_{-T/2}^{+T/2}\left(\frac{1}{2}-\frac{1}{2}\cos2\omega t\right)\,dt\\ &=\frac{1}{T}\left(\frac{1}{2}t\big|_{-T/2}^{T/2}-\frac{1}{4\omega}\sin2\omega t\big|_{-T/2}^{+T/2}\right)\\ &\overset{T\to+\infty,\,\omega=\frac{2\pi}{T}}{=}\frac{1}2-\frac{1}{T}\cdot\frac{T}{8\pi}\sin\left(\frac{4\pi}{T}\cdot\frac{T}{2}\right)\cdot2\\ &=\frac{1}{2} \end{align}

$\endgroup$
7
$\begingroup$

The mean of a function over an interval $T$ is given by,

$$\langle f(t) \rangle = \frac{1}{T}\int_0^T f(t) \, \mathrm{d}t$$

which is roughly the continuous analogue of the arithmetical mean. For $\sin^2 \omega t$, we obtain,

$$\langle \sin^2 \omega t \rangle = \frac{1}{T} \int_0^T \sin^2 \omega t \, \mathrm{d} t = \frac{1}{T} \left( \frac{T}{2} - \frac{\sin(2\omega T)}{4\omega}\right)$$

If we choose $T=2\pi \omega^{-1}$, the result simplifies to,

$$\langle \sin^2 \omega t \rangle = \frac{1}{2}.$$

$\endgroup$
  • 1
    $\begingroup$ Instead of choosing $T = \pi$, usually $\omega = \frac{2\pi}{T}$, and the second term vanishes. (I.e., $T$ is a cycle/period) $\endgroup$ – Minethlos Apr 15 '15 at 22:25
6
$\begingroup$

The average of discrete quantities is:

$$\langle x \rangle = \frac{1}{N}\sum_i x_i$$

The extension to a continuous quantity should seem fairly natural if you think of an integral as a continuous sum:

$$ \langle f(x) \rangle = \frac{1}{b-a} \int_a^b f(x) dx$$

where $[a,b]$ is the interval you want to average over. Then you just need to evaluate this integral. The bounds $(-\infty,\infty)$ make it seem a bit difficult, but for a periodic function the average over one cycle is the same as the average over many cycles.

$\endgroup$
5
$\begingroup$

Another way to see this is by using the fact that $$\sin^2 \omega t=\frac{1-\cos(2\omega t)}{2}$$

and the average value of $\cos(2\omega t)$ is zero.

$\endgroup$
2
$\begingroup$

It can be easily answered graphically by realizing $\displaystyle \sin^2(x) - \frac{1}{2} $ is symmetric about the x-axis. Therefore $$\left\langle \sin^2(x) - \frac{1}{2} \right\rangle = 0$$

Adding $\displaystyle \frac{1}{2} $ to both sides of the equation yields the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy