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My friend went to an interview for a reputed scholarship program and was asked this question. A wave has an equation $a\sin(\omega t-kx)$. Sometimes k surely can become -ve. We know that $k=\frac{2\pi}{\lambda}$. So $\lambda$ is -ve? How can this be?

What he said was that we can write $k=\frac{\omega}{v}$. Since $\omega=\frac{2\pi}{t}$, and t can't be -ve, v should be -ve. So it implies that the wave is in -ve x direction. That is why wavelength has come -ve due to this sign convention. But they didn't agree to it. Even I think the above is correct. Where is the problem then?

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Your equation for the wave is really a vector equation:

$$ \psi({\bf x}, t) = a \sin(\omega t - {\bf k . x}) $$

This tends to be glossed over when students are first taught the equation, and to be fair in 1D the dot product $\bf k.x$ is simply $kx$ or $-kx$ depending on whether $\bf k$ and $\bf x$ point in the same or opposite directions.

Anyhow, $\bf k$ is the wave vector and like all vectors has direction and magnitude. Its magnitude is the wave number $k$, and the wave number is equal to $2\pi/\lambda$. Because $k$ is the magnitude of a vector it is always positive, and therefore $\lambda$ is always positive.

When you have a wave travelling in the $-x$ direction it's not $\lambda$ that changes sign, it's the direction of $\bf k$.

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Thanks for telling that, we were not actually taught this, but you mean that if k is -ve, then $\lambda$ is positive (no matter what)? And -ve k is just the direction of k? –  Rohinb97 Feb 4 at 11:40
    
@Rohinb97: Basically yes, the wavenumber $k$ is always positive. Writing $\omega t - kx$ (NB the non-vector form) means the wavevector points in the positive $x$ direction, while Writing $\omega t + kx$ means the wavevector points in the negative $x$ direction. But note that the wavenumber is positive in both cases. It seems confusing, but it's a result of trying to (over)simplify the vector form of the equation. While vectors are conceptually harder this confusion goes away iof you just use the vector form. –  John Rennie Feb 4 at 12:05
    
So, the above, which my friend said, that $k=\frac{\omega}{v}$, so v is -ve , and that's why $\lambda$ is -ve due to sign convention is wrong? –  Rohinb97 Feb 4 at 13:34

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