Wave Intensity: A^2(z,t) at (15 cm, .5s)

[UrbanXrisis]

the wave intensity is defined as the squared wave amplitude.

$$I(z,t)=A^2 (z,t)$$
$$A(z,t)=a cos [2 \pi (\frac{r}{\lambda}-\frac{t}{T})]$$

the wavelength is 10cm
period is 1/10s
a=1mm


1. what is the instantaneous intensity at r=15cm and t=.5s?

is the solution:
$$A^2(z,t)=(.1 cos [2 \pi (\frac{15}{10}-\frac{.5}{.1})])^2$$

2. what is the average intensity averaged over a long time at that location?

$$\int A^2(15,t)=\int_{0}^{\inf} (.1 cos [2 \pi (\frac{15}{10}-\frac{t}{.1})])^2 dt$$

is that how I would solve #2?

 

[mezarashi]

Integrating like that would give you a total sum. You need to divide it by the time to get the 'average'. A trick would be to integrate from 0 to any T (period of the wave) and divide it by T.