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

1.	A light vertical spring is stretched 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 costant of the spring `k= ("Restoring Force")/("Increase in length")= ("MG")/("Increase in length")( :. F=mg)= (10 xx 10^(-3) xx 9.8)/(2.0 xx 10^(-2))= 4.9 Nm^(-1)` Time period of a spring, `T= 2pisqrt((m)/(k))= 2pisqrt((10 xx 10^(-3))/(4.9))` The frequency , `v= (1)/(T)= (1)/(2pi)SQRT((4.9)/(10^(-2)))= 3.523 HZ` (ii) The amplitude of motion =The distance through which the WEIGHT is further PULLED down = 1.0 cm i.e., A= 1.0 cm.