1. Given SHM equations

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

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

We need the ratio of amplitudes of $y_1$ to $y_2$ in the form $x:1$.

2. Amplitude of $y_1$

The standard SHM form is

$y=A\sin(\omega t+\phi)$

So, from

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

the amplitude is

$A_1=10$

3. Amplitude of $y_2$

Given

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

First, find the amplitude of

$\sin 3\pi t+\sqrt{3}\cos 3\pi t$

For an expression of the form

$a\sin \theta+b\cos \theta$

the amplitude is

$\sqrt{a^2+b^2}$

Here,

$a=1,\qquad b=\sqrt{3}$

So amplitude of the bracketed term is

$\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=2$

Now multiply by 5:

$A_2=5\times 2=10$

4. Ratio of amplitudes

$A_1:A_2=10:10=1:1$

Hence, in the form $x:1$,

$x=1$

5. Comparison with stored answer

Stored correct answer = $1$

Our derived answer = $1$

So the answer agrees.