A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?

A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass

1.	A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer» Solution :When 0.02 kg mass is added, the spring STRETCHES by 7 cm As `MG=kx` `therefore k=(mg)/(x)=(0.02xx10)/(7XX10^(-2))=(20)/(7)NM^(-1)` When 0.02 kg mass is removed, the period of vibration will be `T=2xsqrt((m)/(k))=2pisqrt((0.2)/(20//7))=2pisqrt((7)/(100))=(2pixx2.645)/(10)=1.66s`