QJEEM23Pha480: JEE Main 2023 Physics - Class 11 JEEM23Pha480 Solution: For a periodic motion represented by the equation y = sin ⁡ ω t …

We can write the given equation as:

$$y=\sqrt{(\sin \omega t{)}^2+(\cos \omega t{)}^2}\cos (\omega t-\arctan \frac{\sin \omega t}{\cos \omega t})$$

Using the identity $\sin^2\theta + \cos^2\theta = 1$ , we get:

$$y=\sqrt{1+\sin 2\omega t}\cos (\omega t-\frac{\pi }{4})$$

The amplitude of the motion is the maximum value of $|y|$ , which occurs when $\sin 2\omega t = 1$ , i.e., at $t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega}$ , where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:

$$|y_{\text{max}}|=\sqrt{1+\sin \frac{\pi }{2}}=\sqrt{2}$$

Therefore, the amplitude of the motion is $\sqrt{2}$