NEET2026PhysicsWaves and SoundActual
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 :
Options
- A0.08 rad
- B0.8 rad
- C8 rad
- D0.008 rad
Correct answer
B. 0.8 rad
Step-by-step solution
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