The frequency of oscillation of a mass \(m\) suspended by a spring is \(v_1\). If the length of the spring is cut to half, the same mass oscillates with frequency \(v_2\). The value of \(v_2/v_1\) 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: \(\sqrt{2}\)