For a periodic motion represented by the equation $$y = \sin\omega t + \cos\omega t$$ the amplitude of the motion is

We need to find the amplitude of the periodic motion given by $$y = \sin\omega t + \cos\omega t$$.

We begin by recalling that an expression of the form $$y = A\sin\theta + B\cos\theta$$ can be rewritten as a single sinusoidal function using the identity:
$$A\sin\theta + B\cos\theta = \sqrt{A^2 + B^2}\sin(\theta + \phi)$$ where $$\phi = \tan^{-1}\left(\frac{B}{A}\right)$$.

Next, in our expression $$y = \sin\omega t + \cos\omega t$$, we recognize that $$A = 1$$ (coefficient of $$\sin\omega t$$) and $$B = 1$$ (coefficient of $$\cos\omega t$$).

Substituting these values into the amplitude formula gives the amplitude of the combined motion as $$\text{Amplitude} = \sqrt{A^2 + B^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$$.

Then the phase angle is calculated by $$\phi = \tan^{-1}\left(\frac{B}{A}\right) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4}$$, and therefore the motion can be expressed as $$y = \sqrt{2}\sin\left(\omega t + \frac{\pi}{4}\right)$$.

This confirms that the motion is simple harmonic with amplitude $$\sqrt{2}$$ and angular frequency $$\omega$$.

The correct answer is Option 4: $$\sqrt{2}$$.