The frequency of oscillation of a mass m suspended by a spring is...

The frequency of oscillation of a spring-mass system is given by $v = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass.

The spring constant $k$ is inversely proportional to the length of the spring $L$, so $k \propto \dfrac{1}{L}$.

When the length of the spring is cut to half, the new length is $L' = \dfrac{L}{2}$. The new spring constant becomes $k' = 2k$.

The new frequency of oscillation is $v_2 = \dfrac{1}{2\pi} \sqrt{\dfrac{k'}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{2k}{m}} = \sqrt{2} v_1$.

Therefore, the ratio $\dfrac{v_2}{v_1} = \sqrt{2}$.

Answer: $\sqrt{2}$