NEET 2026 — Physics solution

NEET2026Physics

For a travelling harmonic wave $y (x, t) = 2.0 cos 2 π (10 t - 0.0080 x + 0.35)$ , where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:

A
0.08 $π$ rad
B
0.008 $π$ rad
C
0.8 $π$ rad
D
8 $π$ rad

Solution

From the wave equation, the wavenumber is $k = 2 π \times 0.008$ rad/cm.

For separation $Δ x = 0.5$ m $= 50$ cm,

$Δ ϕ = k Δ x = 2 π \times 0.008 \times 50 = 0.8 π$ rad.

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