Identify the amplitude of the first SHM

The first motion is $x_1 = 5\sin\left(2\pi t + \frac{\pi}{4}\right).$

This is already in the standard SHM form $x = A\sin(\omega t + \phi),$ so its amplitude is $A_1 = 5.$

Rewrite the second SHM in standard form

The second motion is $x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t).$

Use the identity $\sin\theta + \cos\theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right).$

Therefore, $x_2 = 5\sqrt{2}\cdot \sqrt{2}\sin\left(2\pi t + \frac{\pi}{4}\right).$

So, $x_2 = 10\sin\left(2\pi t + \frac{\pi}{4}\right).$

Hence the amplitude of the second motion is $A_2 = 10.$

Find how many times the second amplitude is the first

$\frac{A_2}{A_1} = \frac{10}{5} = 2.$

Final answer

The amplitude of the second motion is $2$ times the amplitude of the first motion.