The frequency of oscillation of a spring-mass system is given by v=2π1mk, 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∝L1.
When the length of the spring is cut to half, the new length is L′=2L. The new spring constant becomes k′=2k.
The new frequency of oscillation is v2=2π1mk′=2π1m2k=2v1.
Therefore, the ratio v1v2=2.
Answer: 2