Three masses 700 ~g , 500 ~g and 400 g are suspended at the end of a spring shown in figure and are in equilibrium. When the 700 ~g mass is removed, the system oscillates with a time period of 3 ~s . If 500 ~g mass is further removed, then it will oscillate with a period of

Step-by-step solution

Time-period of oscillations, T=2 m k where, m= mass of the oscillating system and k= spring constant. Case I 700 ~g mass is removed, then m=500 ~g +400 ~g =900 ~g =0.9 ~g So, T₁=2 0.9 k 3=2 0.9 k ( i ) [ T₁=3 ~s . (given) ] Case II 500 ~g mass is further removed, then m=400 ~g =0.4 ~kg On dividing Eq. (ii) by Eq. (i), we get aligned T₂ 3 & = 2 0.4 k 2 0.9 k T₂ 3 = 0.4 0.9 T₂ 3 = 4 9 T₂ 3 & = 2 3 T₂=2 ~s aligned