Two light spring of force constants `k_(1)` and `k_(2)` and a block of mass m are in one line `AB` on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure. The distance `CD` between the spring is `60cm`. If the block moves along `AB` with a velocity `120 cm//s` in between the springs, calculate the period of oscillation of the block. (take `k_(1) = 1.8 N//m`, `k_(2) = 3.2 N//m`, `m = 200 g`)

Options

A. `1.41 S`
B. `2.81S`
C. `5.64S`
D. `1.92 S`

Answer

B

Explanation

Between `C` and `D` block will move with constant speed of `120cm//s`. Therefore, period of oscillation will be (starting from `C).T = t_(CD) +(T_(2))/(2) +t_(DC) +(T_(1))/(2)`
Here, `T_(1) =2p sqrt((m)/(k_(1)))` and `T_(2) =2p sqrt(m)/(k_(2))`and `t_(CD) =t_(DC) = (60)/(120) =0.5s`
`:. T = 0.5 +(2pi)/(2) sqrt((0.2)/(3.2)) +0.5 +(2pi)/(2) sqrt((0.2)/(1.8))`
`(m = 200g =0.2 kg)T = 2.82 s`