Two simple harmonic motion, are represented by the equations
$$y_1 = 10\sin\left(3\pi t + \frac{\pi}{3}\right)$$; $$y_2 = 5\left(\sin 3\pi t + \sqrt{3}\cos 3\pi t\right)$$
Ratio of amplitude of $$y_1$$ to $$y_2$$ = $$x$$ : 1. The value of $$x$$ is _________.

We begin by identifying the two given simple harmonic motions:

$$y_1 = 10\sin\!\left(3\pi t + \frac{\pi}{3}\right),$$

$$y_2 = 5\left(\sin 3\pi t + \sqrt{3}\cos 3\pi t\right).$$

For any expression of the form $$y = a\sin\omega t + b\cos\omega t,$$ the amplitude is found from the well-known relation

$$\text{Amplitude} = \sqrt{a^{2}+b^{2}}.$$

First we look at $$y_1.$$ Here, the coefficient of the sine term is already a single number, namely $$10,$$ and there is no separate cosine term. Thus the amplitude of $$y_1$$ is directly

$$A_1 = 10.$$

Now we turn to $$y_2.$$ We rewrite it in the separated form to make the two coefficients clear:

$$y_2 = 5\sin 3\pi t + 5\sqrt{3}\cos 3\pi t.$$

So, in the standard notation, we have

$$a = 5, \quad b = 5\sqrt{3}.$$

Applying the amplitude formula, we obtain

$$A_2 = \sqrt{a^{2}+b^{2}} = \sqrt{\left(5\right)^{2} + \left(5\sqrt{3}\right)^{2}} = \sqrt{25 + 25\cdot 3} = \sqrt{25 + 75} = \sqrt{100} = 10.$$

Thus the amplitudes come out equal:

$$A_1 = 10, \qquad A_2 = 10.$$

The required ratio of the amplitudes is therefore

$$\frac{A_1}{A_2} = \frac{10}{10} = 1.$$

This means

$$x : 1 = 1 : 1,$$

so

$$x = 1.$$

So, the answer is $$1$$.