Q183 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
A progressive wave is given by $Y = 6 \sin (3t - 2x)$ m. On this wave, how far away are the two points having phase difference of $45^\circ$ ?
- A$\dfrac{\pi}{16}\text{m}$
- Correct answerB$\dfrac{\pi}{8}\text{m}$
- C$\dfrac{\pi}{4}\text{m}$
- D$\dfrac{\pi}{2}\text{m}$
Solution
The given equation of the progressive wave is $Y = 6 \sin (3t - 2x)$.
Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$, we get the wave number $k = 2$.
The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given $\Delta \phi = 45^\circ = \dfrac{\pi}{4}$ rad.
Substituting the values, we get:
$\dfrac{\pi}{4} = 2 \Delta x$
$\Delta x = \dfrac{\pi}{8}$ m.
Answer: $\dfrac{\pi}{8}\text{m}$
Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$, we get the wave number $k = 2$.
The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given $\Delta \phi = 45^\circ = \dfrac{\pi}{4}$ rad.
Substituting the values, we get:
$\dfrac{\pi}{4} = 2 \Delta x$
$\Delta x = \dfrac{\pi}{8}$ m.
Answer: $\dfrac{\pi}{8}\text{m}$
Hence, the correct answer is option: 'B'.