Q115 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
The minimum phase difference between two simple harmonic motions is
$x_1 = \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t$
$x_2 = \sin \omega t + \cos \omega t$ $\left[ \sin \dfrac{\pi}{4} = \cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} \right]$
$x_1 = \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t$
$x_2 = \sin \omega t + \cos \omega t$ $\left[ \sin \dfrac{\pi}{4} = \cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} \right]$
- Correct answerAzero
- B$\dfrac{\pi^C}{3}$
- C$\dfrac{\pi^C}{4}$
- D$\dfrac{\pi^C}{5}$
Solution
The given equations of simple harmonic motion are:
$x_1 = \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t$
$x_1 = \sin \omega t \cos \dfrac{\pi}{4} + \cos \omega t \sin \dfrac{\pi}{4} = \sin\left(\omega t + \dfrac{\pi}{4}\right)$
$x_2 = \sin \omega t + \cos \omega t$
$x_2 = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t \right) = \sqrt{2} \sin\left(\omega t + \dfrac{\pi}{4}\right)$
The phase of $x_1$ is $\phi_1 = \dfrac{\pi}{4}$ and the phase of $x_2$ is $\phi_2 = \dfrac{\pi}{4}$.
Phase difference $\Delta \phi = \phi_1 - \phi_2 = \dfrac{\pi}{4} - \dfrac{\pi}{4} = 0$.
$x_1 = \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t$
$x_1 = \sin \omega t \cos \dfrac{\pi}{4} + \cos \omega t \sin \dfrac{\pi}{4} = \sin\left(\omega t + \dfrac{\pi}{4}\right)$
$x_2 = \sin \omega t + \cos \omega t$
$x_2 = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t \right) = \sqrt{2} \sin\left(\omega t + \dfrac{\pi}{4}\right)$
The phase of $x_1$ is $\phi_1 = \dfrac{\pi}{4}$ and the phase of $x_2$ is $\phi_2 = \dfrac{\pi}{4}$.
Phase difference $\Delta \phi = \phi_1 - \phi_2 = \dfrac{\pi}{4} - \dfrac{\pi}{4} = 0$.
Hence, the correct answer is option: 'A'.