A mass m is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period T . If the mass is increased by m ₀ , the periodic time becomes 5 ~T 4 . The ratio m ₀ M is

Step-by-step solution

The period of oscillation for a mass-spring system is given by T = 2 m k . When the mass increases to m + m₀ , the new period becomes T' = 5T 4 . Taking the ratio of periods: T' T = m + m₀ m Substituting T' = 5T 4 yields: 5 4 = m + m₀ m Squaring both sides: ( 5 4 )^2 = m + m₀ m Which simplifies to: 25 16 = 1 + m₀ m Solving for the ratio: m₀ m = 25 16 - 1 = 9 16 The ratio of additional mass to original mass is 9 16 .