Physics · Waves · NEET
For a wave y = a sin(k x - w t), two points at positions x1 and x2 (at the same instant) differ only in the k x term. So phase difference delta-phi = k times (x2 - x1) = k times delta-x. Since k = 2 pi / lambda, you get delta-phi = (2 pi / lambda) times delta-x. Here delta-x is called the path difference. This is a spatial phase difference (same time, different places). Keep delta-x and lambda in the same unit before dividing.
They are directly proportional: phase difference / (2 pi) = path difference / lambda. Rearranged, delta-phi = (2 pi / lambda) times delta-x, and delta-x = (lambda / 2 pi) times delta-phi. A path difference of lambda gives phase 2 pi; lambda/2 gives pi; lambda/4 gives pi/2. Remember this ratio for NEET interference and beats questions too.
When delta-x = lambda, delta-phi = (2 pi / lambda) times lambda = 2 pi radians. Physically the two points are doing exactly the same thing at every instant (both up together, both down together). This is why points separated by any whole number of wavelengths are said to be 'in phase.'
Two points are in phase when their path difference is a whole number of wavelengths: delta-x = n times lambda, so delta-phi = 2 n pi (they move identically). They are exactly out of phase (opposite motion) when delta-x = (n + 1/2) lambda, so delta-phi = (2 n + 1) pi. For NEET, 'in phase' means displacement and velocity match; 'out of phase' means one goes up while the other goes down.
No. For a single travelling wave, the term w t is the same for both points at any given instant, so it cancels out. Only the k x term survives, giving delta-phi = k times delta-x, which is fixed by their separation. Time only shifts the whole pattern; the gap between the two points stays constant. (Do not confuse this with the phase difference of one point between two different times, which is w times delta-t.)
Use one consistent unit for both. The mistake NEET punishes is mixing units, for example lambda in cm and delta-x in m. In the 2026 PYQ the equation had x in cm, so delta-x = 0.5 m had to be converted to 50 cm first. Convert, then apply delta-phi = k times delta-x.
For a travelling harmonic wave y(x, t) = 2.0 cos[2 pi(10t - 0.0080x + 0.35)], where x and y are in cm and t in s, the phase difference between the oscillatory motion of two points separated by 0.5 m is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Spatial phase difference is between two different points at the same time: delta-phi = k times delta-x. Temporal phase difference is for one point at two different times: delta-phi = w times delta-t. This page is about the spatial one (two points on the wave).
Yes. If delta-x is more than one wavelength, delta-phi exceeds 2 pi. In the 2026 PYQ, delta-x = 50 cm while lambda = 1 / 0.0080 = 125 cm, so delta-x is less than lambda and delta-phi = 0.8 pi (under 2 pi). For larger separations you can subtract whole multiples of 2 pi to find the 'effective' phase.
Use lambda = 2 pi / k. Once you have lambda, delta-phi = (2 pi / lambda) times delta-x. Or skip lambda entirely and use delta-phi = k times delta-x directly, since k already equals 2 pi / lambda.
At a single instant, both points share the same time t, so w t is identical for them and cancels when you subtract. Only their positions differ, so only the k x term contributes to delta-phi.
