A ball of mass m = 1 kg is hung from a spring of constant 400 N / m and another ball of mass 2 m is hanging from it as shown with the help of a thread. The system is in equilibrium. Now the thread is

Question

A ball of mass m = 1 kg is hung from a spring of constant 400 N / m and another ball of mass 2 m is hanging from it as shown with the help of a thread. The system is in equilibrium. Now the thread is burned. As a result, ball of mass m starts moving upwards. Eventually it will stop under the influence of gravity and spring. In between, there is a point at which its speed is maximum. What is that maximum speed (in m/s) ?

TestHub · Accepted Answer

Initial stretch: x 1 ​ = k ( m + 2 m ) g ​ = 400 30 ​ = 0.075 m . After the thread burns, the 1 kg ball oscillates about its own equilibrium x 2 ​ = k m g ​ = 0.025 m , so the amplitude is 0.05 m. v ma x ​ = ω A = 1 400 ​ ​ × 0.05 = 1 m/s Ans. 1

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