Calculating Temporal Average of Sin^2 Function

[fluidistic]

Homework Statement

The temporal average of a function f(t) over an interval T is given by $$\langle f(t) \rangle=\frac{1}{T} \int _{t}^{t+T} f(t') dt'$$.
Let $$\tau =\frac{2 \pi}{\omega}$$. Calulate the temporal average of $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}$$, when $$T=\tau$$ and $$T >> \tau$$.

Homework Equations

Already given in the problem description.

The Attempt at a Solution

I don't understand the problem. Aren't they asking me to prove that $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}$$ for $$T=\tau$$ and to calculate $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle$$ for $$T>> \tau$$? If so, it's really not clear. If not, could you please explain with a bit more details? I can't start the problem since I don't understand it.

 

[jdwood983]

Looks to me like the question is asking you to show that

$$\left\langle \sin^2\left[\vec{k}\cdot\vec{r}-\omega t\right]\right\rangle=\frac{1}{2}$$

for both $T=\tau$ and $T\gg\tau$.

 

[fluidistic]

Looks to me like the question is asking you to show that

$$\left\langle \sin^2\left[\vec{k}\cdot\vec{r}-\omega t\right]\right\rangle=\frac{1}{2}$$

for both $T=\tau$ and $T\gg\tau$.

Ok thanks a lot. I'll give some tries.

 

[gabbagabbahey]

The first case, $T=\tau$ can be easily done with a useful trick...if you graph sin^2(u) and cos^2(u) over a full period, you should see that the area under them is equal and so

$$\int_{u_0}^{u_0+2\pi}\sin^2(u)du=\int_{u_0}^{u_0+2\pi}\cos^2(u)du=\frac{1}{2}\left(\int_{u_0}^{u_0+2\pi}\sin^2(u)du+\int_{u_0}^{u_0+2\pi}\cos^2(u)du\right)=\frac{1}{2}\int_{u_0}^{u_0+2\pi}du=\pi$$

For the second case, I think you will want to use the double angle formula, and then expand your result in a Taylor series.

 

[paranormal]

As a scientist, my response would be:

The problem statement is asking for the temporal average of the function \sin^2(\vec{k}\cdot\vec{r}-\omega t) over an interval T, where T is equal to the period of the function, \tau, and also for T much larger than \tau. This means that we are interested in finding the average value of the function over one full cycle (T=\tau) and over multiple cycles (T>>\tau).

To solve this problem, we can use the definition of temporal average provided in the problem statement, which involves taking the integral of the function over the given interval and dividing by the length of the interval. For T=\tau, we can rewrite the function as \sin^2(\vec{k}\cdot\vec{r}-\omega t) = \sin^2(\vec{k}\cdot\vec{r}-2\pi t) since \tau = \frac{2\pi}{\omega}. This function has a period of 2\pi, meaning that over one full cycle, the function will have the same value at t=0 and t=2\pi. Therefore, the integral of the function over the interval \tau will be equal to the integral of the function over one full cycle, which is 2\pi. Dividing this by the length of the interval, \tau, we get \frac{2\pi}{\tau} = \frac{2\pi}{\frac{2\pi}{\omega}} = \omega. Since the function \sin^2(\vec{k}\cdot\vec{r}-\omega t) has a maximum value of 1 and a minimum value of 0, the average value over one full cycle will be \frac{1}{2}, which is what we were asked to prove.

For T>>\tau, we can use the same approach to calculate the temporal average. The function \sin^2(\vec{k}\cdot\vec{r}-\omega t) will have multiple cycles within the interval T, and therefore, the integral of the function over T will be equal to the sum of the integrals over each individual cycle. Since each cycle has an average value of \frac{1}{2}, the average value over multiple cycles will also be \frac{1}{2}. This can be mathematically expressed as \langle \sin ^2 (\