Intensity and Superposition of waves

[Abhishekdas]

Intensity and Superposition of waves...

Homework Statement

Incident wave y=Asin(ax + bt + pi/2) is reflected by an obstacle at x=0 which reudces intensity of reflected wave by 36%. Due to superposition a resulting wave consist of standing wave and traveling wave given by y= -1.6 sinax.sinbt + cAcos(bt+ax)

Find amplitude of reflected wave, value of c and position of second antinode...

Homework Equations

Intensity is proportional to amplitude square...

The Attempt at a Solution

Now how are they getting this type of a n equation and what do they mena by a wave consisting of a standing as well as a traveling wave at the same time? i want help in finding out the value of these constants...All i got is amplitude of the reflected wave is 3/5 times the incident waves' amplitude...(By Intensity is proportional to amplitude square...)...Thats it... please help...

 

[jambaugh]

This is a problem involving trigonometric identities.

I suggest you use the product formula for sin(ax)sin(bt) (you can use complex exponentials if you prefer) to write it as a sum of sines or cosines of (ax+bt) and (ax-bt).


Let's see... the identity is:
\sin(\alpha)\sin(\beta) = \frac{e^{i\alpha}-e^{-i\alpha}}{2i}\frac{e^{i\abeta}-e^{-i\beta}}{2i} =
= \frac{1}{-4} (e^{i(\alpha+\beta)}+e^{-i(\alpha+\beta)} - e^{i(\alpha-\beta)}-e^{-i(\alpha - \beta)})
= \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha+\beta)]

 

[Abhishekdas]

Hi jambaugh...thanks for your reply ...
I got the identity but how does it apply here? I am afraid i still didnt get much of a clue about how this equation comes...