When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period?(g=10ms-1)

When 0.5 kg is suspended, extension of the spring is 0.2 m. Since in equilibrium mg=kxWhen 0.5 kg is suspended, extension of the spring is 0.2 m.  in equilibrium mg=kx 0.5×10=k×0.2 spring constant =k=25 N/m  Time period of spring block system when 0.25 kg mass is attached time period =T=2πmK substitute m, K values T=2π0.2525 T=2π252500 T=2×3.1410 T=0.628 seconds