A tolley of mass 3.0kg is connected to two identical spring each of force constant `600Nm^(-1)` as shown in figure. If the trolley is displaced from its equilibrium position by 5.0 cm and released, what is
(a) the period of ensuing oscillation?
(b) the maximum speed of trolley?
(c) How much is the total energy dissipated as heat by the time the trolley comes to rest due to damping forces?

Explanation

Here, `m=3.0kg. `
`k_(1)=k_(2)=600Nm^(-1), a=5.0cm =0.05m`
When the mass is displaced a little to one side, one spring gets compressed and another is elongated. Due to which the combination of spring works as parallel combination of springs. Here, effective spring factore K will be given by
`K=k_(1)+k_(2)=600+600=1200Nm^(-1)`
(a) `T=2pisqrt((m)/(K))=2xx3.14sqrt((m)/(1200))=3.14s`
(b) `V_(max)=aomega =asqrt((K)/(m))`
`=0.05sqrt((1200)/(3))=1ms^(-1)`
(c) Total energy, `E=(1)/(2)Ka^(2)`
`=(1)/(2)xx1200xx(0.05)^(2)=1.5J`