For a travelling harmonic wave y(x, t) = 2.0\ \ 2 (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 :

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 0.8\ rad

The given equation of the travelling harmonic wave is y(x, t) = 2.0 2 (10 t - 0.0080 x + 0.35). Comparing this with the standard wave equation y(x, t) = A ( t - kx + _0), we get the wave number k = 2 0.0080 rad/cm. The distance between the two points is given as x = 0.5 m = 50 cm. The phase difference between two points separated by a distance x is given by = k x. Substituting the values, we get: = (2 0.0080) 50 = 2 0.4 = 0.8 rad. Answer: 0.8\ rad

NEET Physics Question (2026) — Solution

Question

Options

Answer

Step-by-step solution