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Calculate the time period of the oscillation of a particle of mass m moving in the potential defined as U(x)={{:((1)/(2) kx^(2)",", x lt 0),(mgx"," , g gt 0):} |
Answer» Solution : In the problem oscillation of particle into 2 cases via `V lt 0` (consider it as SHM where time period is considered as `t_(1))` and another one as `x GT 0` (consider it as motion under gravity, where time period is `t_(2))` Note : We want to find total time period, which will be `T=t_(1)+t_(2)` According to conservation of energy `(1)/(2)kx^(2)=(1)/(2) mv^(2)` `:.v^(2)=(kx^(2))/(m)` `rArr v= sqrt((kx^(2))/(m)) ""...(1)` We KNOW that, kinetic energy (E) } `(1)/(2) mv^(2)` `v=sqrt((2E)/(m)) ""....(2)` In SHM (case : 1) `T = 2pi sqrt((m)/(k))` Here , `t_(1)=(T)/(2)` `= (2pi sqrt((m)/(k)))/(2)` `= pi sqrt((m)/(k)) ""...(3)` In case (motion under gravity) `t_(2)=(2V)/(g)` substituting equation (2) have we get, `t_(2)=(2)/(g) sqrt((2E)/(m))` `rArr 2sqrt((2E)/(mg^(2))) ""....(4)` Adding equation (3) & (4) Time period of oscillation, `T=t_(1)+t_(2)` `= pi sqrt((m)/(k))+2 sqrt((2E)/(mg^(2)))` |
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