A light spring is suspended with mass ' m₁ ' at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is T second. When a mass ' m₂ ' is added to ' m₁ ' and the system is made to oscillate the period is found to be 3 2 T. The ratio ( m₁ m₂ ) is

Step-by-step solution

The time period of a spring-mass system is given by T = 2 m k , where m is the mass and k is the spring constant. For the initial case with mass m₁ : T = 2 m₁ k When mass m₂ is added, the total mass becomes m₁ + m₂ and the new time period is 3 2 T : 3 2 T = 2 m₁ + m₂ k Dividing the second equation by the first equation: 3 2 = m₁ + m₂ m₁ Squaring both sides: 9 4 = m₁ + m₂ m₁ 9 4 = 1 + m₂ m₁ m₂ m₁ = 9 4 - 1 = 5 4 Taking the reciprocal to find the required ratio: m₁ m₂ = 4 5 The ratio is 4 : 5 . Answer: 4 : 5