Three masses 100g, 300g and 500g are suspended at the end of a spring as shown in figure and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 3 second. When the 300 g mass is also removed, it will oscillate with a period of

Step-by-step solution

Let the spring constant be k . When the 500 g mass is removed, the remaining mass attached to the spring is m₁ = 100 g + 300 g = 400 g . The time period of oscillation is given by T₁ = 2 m₁ k . Given T₁ = 3 s , we have 3 = 2 400 k . When the 300 g mass is also removed, the remaining mass is m₂ = 100 g . The new time period is T₂ = 2 m₂ k = 2 100 k . Taking the ratio of the two time periods: T₂ T₁ = m₂ m₁ = 100 400 = 1 2 Therefore, T₂ = T₁ 2 = 3 2 = 1.5 s . Answer: 1.5 s