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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?

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  • $\begingroup$ Which textbook? Which page? $\endgroup$
    – Qmechanic
    Oct 8 at 5:32
  • $\begingroup$ Concepts of Physics part 1. Page 322 $\endgroup$
    – Mr. Wayne
    Oct 8 at 5:41
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    $\begingroup$ 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. :-) $\endgroup$
    – StephenG
    Oct 8 at 13:03
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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 $\frac{A}{2}$.

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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 $A/2$.

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