Two simple harmonic motions are represented by the equations $$x_1 = 5\sin\left(2\pi t + \frac{\pi}{4}\right)$$ and $$x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t)$$. The amplitude of the second motion is _________ times the amplitude in the first motion.

We are given two simple harmonic motions. The first displacement is written as $$x_1 = 5 \sin\left( 2\pi t + \dfrac{\pi}{4}\right)$$. In any expression of the form $$x = A \sin(\omega t + \phi)$$, the coefficient $$A$$ in front of the sine (or cosine) function is the amplitude. Hence, for the first motion we directly read off the amplitude

$$A_1 = 5.$$

Now we examine the second displacement, $$x_2 = 5\sqrt{2}\,(\sin 2\pi t + \cos 2\pi t).$$ We first need the amplitude of the bracketed term $$\bigl(\sin 2\pi t + \cos 2\pi t\bigr)$$ itself. A sum of sine and cosine with the same angular frequency can be combined into a single sine (or cosine) term using the identity

$$\sin \theta + \cos \theta \;=\; \sqrt{2}\,\sin\!\left(\theta + \dfrac{\pi}{4}\right),$$

because

$$\sqrt{2}\,\sin\!\left(\theta + \dfrac{\pi}{4}\right) = \sqrt{2}\,\Bigl[\sin\theta \cos\dfrac{\pi}{4} + \cos\theta \sin\dfrac{\pi}{4}\Bigr]$$

and since $$\cos\dfrac{\pi}{4} = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2},$$ we get

$$\sqrt{2}\,\Bigl[\sin\theta \,\dfrac{\sqrt{2}}{2} + \cos\theta \,\dfrac{\sqrt{2}}{2}\Bigr] = \sin\theta + \cos\theta.$$

Thus we may rewrite the bracketed term as

$$\sin 2\pi t + \cos 2\pi t = \sqrt{2}\,\sin\!\left(2\pi t + \dfrac{\pi}{4}\right).$$

This shows that the amplitude associated with the bracketed term alone is $$\sqrt{2}.$$ Multiplying by the outside factor $$5\sqrt{2}$$, the overall amplitude of $$x_2$$ becomes

$$A_2 = 5\sqrt{2}\times \sqrt{2} = 5 \times 2 = 10.$$

To find how many times the second amplitude exceeds the first, we take their ratio:

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

So, the answer is $$2$$.