# Amplitude of a wave [closed]

The wave equation $$y= A \sin^2(kx-\omega t)$$ should have an amplitude $$A$$. But in textbook, it is given that amplitude is $$A/2$$. Can anyone explain?

• Which textbook? Which page? Oct 8 at 5:32
• Concepts of Physics part 1. Page 322 Oct 8 at 5:41
• As a general rule please give the author name(s) to physics books as they tend to have very similar (even identical titles) - e.g. "Introduction to ...". No one said physicists were imagnitative. :-) Oct 8 at 13:03

Note that by using the simple trigonometric identity $$\cos(2x)=1-2\sin^2x$$ then
$$\rightarrow \sin^2x=\frac{1-\cos 2x}{2}$$
Therefore we can write the equation $$y= A \sin^2(kx-mt)\tag1$$ as
$$y= \frac{A}{2}[1−\cos(2kx−2mt)]$$ which means that equation (1) actually represents a wave with amplitude $$\frac{A}{2}$$.
The sine squared function ranges between $$0$$ and $$1$$.
The given function ranges between $$y=0$$ and $$y=A$$ and is symmetrical about $$y=A/2$$. So you can think of it as an oscillating function of amplitude $$A/2$$ with an offset of $$A/2$$.