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In the figure shown the spring is relaxed and mass m is attached to the spring. The spring is compressed by 2A and released at t = 0. Mass m collides with the wall and loses two third of its kinetic energy and returns. Starting from t = 0, find the taken by it to come back to rest again (instant at which spring is given undermaximum compression). Take sqrt(m)/(k) = (12)/(pi) |
Answer»  The motion STARTS from position `A` the time taken from `A` to `W_(2) (t_(1)) = (T)/(4) + (T)/(12)` Before collision the energy of the block just before collision `K_(i) = (1)/(2)K(2A)^(2) - (1)/(2) KA^(2)` & just after collision `K_(1) = (K_(f))/(3)` (given) `= (1)/(2)KA^(2)` Now during motion after collision, the energy is again conserved Hence, `K_(1) + (1)/(2) KA^(2) = (1)/(2)KA^('2)` `A' =` maximum compresion after collision `rArr = Asqrt(2)` IE. Now motion has amlitude `Asqrt(2)` Now time taken by block from `W_(2)` to positon `B = (T)/(4) + (T)/(8)` `:.` total time taken `= t_(1) + t_(2) = (T)/(4) + (T)/(12) + (T)/(4) + (T)/(8) = (17)/(24)T = (17pi)/(12)sqrt((m)/(K)) = 17sec {sqrt((m)/(k)) = (12)/(pi)}` |
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