A wave of wavelength $\lambda$ corresponds to $2\pi$ radians. Is this true for all waves? [closed]

Question

A wave of wavelength $\lambda$ corresponds to $2\pi$ radians. Is this true for all waves? Does the $\sin 2x$ wave of wavelength $\lambda$ also corresponds to $2\pi$ radians? I see in graph that it corresponds to $\pi$ radians only…

Answer 1

Yes, the wavelength $\lambda$ always corresponds to a phase difference of $2\pi$ radians, for all waves.

A wave in one dimension can be described by the function $$y=A\sin(2\pi\frac{x}{\lambda})$$ The expression inside the brackets is the phase of the wave $\phi=2\pi\frac{x}{\lambda}$. This tells you the stage of the displacement $y$ in the cycle of the wave, with $2\pi$ being a full cycle - ie the displacement has come back to its initial value. See What is a phase of a wave and a phase difference?

So after moving a distance $x=\lambda$ from the origin, where displacement is zero and the phase $\phi=0$, the displacement of the wave comes back to zero and the phase of the wave becomes $\phi=2\pi\frac{\lambda}{\lambda}=2\pi$.

The trouble is that what is plotted in your graph is not $y$ vs $x$ nor even $y$ vs $\phi$. It is not clear what is being represented, other than the mathematical function $y=\sin(2x)$ in which $x$ appears to be an angle not a distance, in which case it cannot equal the wavelength.

If we did interpret the horizontal axis as distance, measured eg in units of metres, then comparing with the above wave formula we get $$y=\sin(2x)=1.\sin(2\pi\frac{x}{\pi})$$ So we see that the wavelength of this wave is indeed $\pi$ metres. But this is a wavelength in metres, not a phase angle in radians.

No doubt this is quite confusing, but that comes from someone mixing up mathematical functions like $y=\sin(2x)$ with the physics of waves.