For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $$1$$ kg, the angular frequency is $$\omega_1$$. When the mass block is $$2$$ kg the angular frequency is $$\omega_2$$. The ratio $$\frac{\omega_2}{\omega_1}$$ is:

For a mass-spring system, the angular frequency of simple harmonic motion is:

$$\omega = \sqrt{\frac{k}{m}}$$

where $$k$$ is the spring constant and $$m$$ is the mass of the block.

When $$m_1 = 1$$ kg, we get $$\omega_1 = \sqrt{\frac{k}{1}} = \sqrt{k}$$.

When $$m_2 = 2$$ kg, we get $$\omega_2 = \sqrt{\frac{k}{2}}$$.

Now taking the ratio:

$$\frac{\omega_2}{\omega_1} = \frac{\sqrt{k/2}}{\sqrt{k}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

So, the answer is $$\frac{1}{\sqrt{2}}$$.