For a periodic motion represented by the equation

$y=\sin \omega \mathrm{t}+\cos \omega \mathrm{t}$

the amplitude of the motion is

<p>We can write the given equation as:</p> <p>$$y = \sqrt{(\sin\omega t)^2 + (\cos\omega t)^2} \cos\left(\omega t - \arctan\frac{\sin\omega t}{\cos\omega t}\right)$$</p> <p>Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:</p> <p>$y = \sqrt{1 + \sin 2\omega t} \cos\left(\omega t - \frac{\pi}{4}\right)$</p> <p>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:</p> <p>$|y_{\text{max}}| = \sqrt{1 + \sin \frac{\pi}{2}} = \sqrt{2}$</p> <p>Therefore, the amplitude of the motion is $\sqrt{2}$</p>