The frequency of oscillation of a mass m suspended by a spring is v₁ . If the length of the spring is cut to half, the same mass oscillates with frequency v₂ . The value of v₂/v₁ is ________.

Step-by-step solution

The frequency of oscillation of a spring-mass system is given by v = 1 2 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 1 L . When the length of the spring is cut to half, the new length is L' = L 2 . The new spring constant becomes k' = 2k . The new frequency of oscillation is v₂ = 1 2 k' m = 1 2 2k m = 2 v₁ . Therefore, the ratio v₂ v₁ = 2 . Answer: 2