Q191 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
A simple harmonic progressive wave is represented as $Y = 0.04 \sin\pi (2t - 0.02x)$ m. At a given instant of time, the phase difference between two particles 10 m apart is (in radian)
- A$\pi$
- B$\dfrac{\pi}{2}$
- C$\dfrac{\pi}{4}$
- Correct answerD$\dfrac{\pi}{5}$
Solution
The given equation of the wave is $Y = 0.04 \sin(\pi(2t - 0.02x))$.
The phase of the wave at position $x$ and time $t$ is $\phi = 2\pi t - 0.02\pi x$.
Comparing this with the standard wave equation $Y = A \sin(\omega t - kx)$, the propagation constant is $k = 0.02\pi$.
At a given instant of time, the phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given the distance between the particles is $\Delta x = 10$ m.
Substituting the values, we get $\Delta \phi = 0.02\pi (10) = 0.2\pi = \dfrac{\pi}{5}$ radian.
Answer: $\dfrac{\pi}{5}$
The phase of the wave at position $x$ and time $t$ is $\phi = 2\pi t - 0.02\pi x$.
Comparing this with the standard wave equation $Y = A \sin(\omega t - kx)$, the propagation constant is $k = 0.02\pi$.
At a given instant of time, the phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given the distance between the particles is $\Delta x = 10$ m.
Substituting the values, we get $\Delta \phi = 0.02\pi (10) = 0.2\pi = \dfrac{\pi}{5}$ radian.
Answer: $\dfrac{\pi}{5}$
Hence, the correct answer is option: 'D'.