A spring is elongated by 2 cm due to a 80 g mass attached to it. Another body of mass 600g is attached to the end of the spring and it is displaced by 8 cm from its equilibrium position. Calculate the energy of the system in this position. Considering the principle of conservation of energy, determine the velocity of the body when it is at a distance of 4 cm.

A spring is elongated by 2 cm due to a 80 g mass attached to it. Anoth

1.	A spring is elongated by 2 cm due to a 80 g mass attached to it. Another body of mass 600g is attached to the end of the spring and it is displaced by 8 cm from its equilibrium position. Calculate the energy of the system in this position. Considering the principle of conservation of energy, determine the velocity of the body when it is at a distance of 4 cm.
Answer» SOLUTION :FORCE constant of the spring, `k=(80xx980)/2=40xx980"dyn"cm^(-1)` Mass, m = 600 g , amplitude, A = 8 cm. Total energy, E = MAXIMUM potential energy = potential energy at the ends of the path of motion = `1/2kA^(2)=1/2xx40xx980xx(8)^(2)=1254400"erg"=0.12544J`. Even for x = 4 cm, the total energy REMAINS unchanged. If v is the velocity at this POSITION, then `1/2mv^(2)+1/2kx^(2)=1/2xx40xx980xx64` or, `1/2mv^(2)=1/2xx40xx980xx64-1/2xx40xx980xx4^(2)` = `1/2xx40xx980xx(64-16)` = `1/2xx40xx980xx48` or, `v^(2)=(40xx980xx48)/m=(40xx980xx48)/600=4xx49xx16` or, `v=sqrt(4xx49xx16)=2xx7xx4=56cms^(-1)`.