A light vertical spring is streched by 2 cm when a weight of 10 kg is attached to its free end. The weight is further pulled down by 1.0 cm and released. Find its time period and amplitude?

A light vertical spring is streched by 2 cm when a weight of 10 kg is

1.	A light vertical spring is streched by 2 cm when a weight of 10 kg is attached to its free end. The weight is further pulled down by 1.0 cm and released. Find its time period and amplitude?
Answer» Solution :(i) Force constant of the SPRING `K=("Restoring Force")/("INCREASE in length")=(mg)/("Increase in length")( :. F=mg)` `=(10xx10^(-3)xx9.8)/(2.0xx10^(-2))=4.9Nm^(-1)` TIME period of a spring `T=2pisqrt(m/k)=2pisqrt((10xx10^(-3))/4.9)` the frequency `v=1/T=1/(2pi)sqrt((4.9)/(10^(-2)))=3.523Hz` (II) The amplitude of motion= The distance through which the weight is further pulled down =1.0 cm i.e. A=1.0cm