# How can amplitude affect wavelength $\lambda$? [closed]

This was a question published by Cambridge International for an AS Level question last year:

For (II), the mark scheme says that $$v=0.5\lambda × f$$, thus giving double the wavelength as supposed the one shown on the graph. If you try visualizing it, you would see that that the graph for $$A$$ (not $$A^2$$) against $$x$$ would have a wave with lower amplitude. Intuitively you would also modify $$\lambda$$ to half the length so that it looks to scale.

But this is odd when we know the fact that $$A$$ is independent of both $$f$$ and $$\lambda$$. In this case, shouldn't the detector simply square the amplitude ($$A$$) without altering any values on $$x/\mathrm{cm}$$? What am I missing here?

## closed as off-topic by G. Smith, John Rennie, ZeroTheHero, Jon Custer, Cosmas ZachosAug 27 at 13:44

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This has nothing to do with the value of the amplitude or some relationship between amplitude and wavelength. When you square the amplitude the minima (negative values) become maxima (positive). So on the graph above you have twice as many maxima. So according to the graph, the wavelength should be 50 cm (and not 25 cm).

• I've corrected my question. Thanks – Kevin N Aug 25 at 9:44

The graph of $$A$$ has positive and negative values. The graph of $$A^2$$ is always positive, and has twice as many peaks.

Or if you prefer trig to drawing graphs, $$\cos^2 x = \frac 1 2 (\cos 2x + 1)$$.

(I) At x=25 seconds and x=50 seconds the particles of air are 180 degrees out of phase, with one of the positions representing a maximal compression and one representing a maximal rarefaction.

(II) This makes the wavelength=50 cm. Each successive maximum on the $$A^2$$ squared graph represents alternately a maximum and a minimum (or maximal negative) value on a graph of A since the squared positive and squared negative displacements will both become maximums on the $$A^2$$ graph. So the key is that wavelength of the sound wave is 50 cm making the frequency $$33,000cmsec^-1/50cm$$ or 660 Hz.

A key bit of information that you are given in this question is that you are dealing with a stationary wave and so you will need to recall some properties of such waves.

Using the graph that you are given identify the positions of the nodes and antinodes.