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The block of mass m_(1) and m_(2) are connected with a spring of netural length l and spring constant k. The systeam is lying on a smoth horizontal surface. Initially spring is compressed by x_(0) as shown in figure. Show that the two blocks will perform SHM about their equilibrium position. Also (a) find the time period, (b) find amplitude of each block and (c) length of spring as a funcation of time. |
Answer» SOLUTION :(a) Hence both the blocks willbe in equilibrium at the same time when spring is in its natural length. Let `EP_(1)` and `EP_(2)` be equilibrium positions of block `A` and `B` as shown in figure.  Let at any time during oscillations, blocks are at a DISTANCE of `x_(1)` and `x_(2)` from their equilibrium positions. As no external force is acting on the spring on the spring block systeam `:. (m_(1) + m_(2))Deltax_(cm) = m_(1)x_(1) - m_(2) x_(2) = 0` or`m_(1)x_(1) - m_(2) x_(2)` for `1st` particle, force equation can be written as `k(x_(1) + x_(2)) = -m_(1)(d^(2)x_(1))/(dt^(2))` or `k(x_(1) + (m_(1))/(m_(2))x_(1)) = - m_(1)a_(1)` or, `a_(1) = -(k(m_(1) + m_(2)))/(m_(1)m_(2))x_(1) :. omega^(2) = (k(m_(1) + m_(2)))/(m_(1)m_(2))` Hence, `T = 2pisqrt((m_(1)m_(2))/(k(m_(1) + m_(2)))) = 2psqrt((mu)/(K))` where `mu = '(m_(1)m_(2))/((m_(1) + m_(2))` which is known as reduced MASS Ans (a) Similarly time PERIOD of `2`nd particle can be found. Both will be having the same time period. (b) Let the amplitude of blocks be `A_(1)` and `A_(2)`. `m_(1)A_(1) = m_(2)A_(2)` By energy conservation, `(1)/(2)k(A_(1) + A_(2))^(2) = (1)/(2) kx_(0)^(2)` or, `A_(1) + A_(2) = x_(0)` or, `A_(1) + A_(2) = x_(0)` or, `A_(1) + (m_(1))/(m_(2))A_(1) = x_(0)` or, `A_(1) = (m_(2)x_(0))/(m_(1) + m_(2))` similarly, `A_(2) = (m_(1)x_(0))/(m_(1) + m_(2))` (c) Consider equilibrium position of `1`st particle as origin, i.e. `x = 0`. `x` co-ordinate of particles can be written as `x_(1) A_(1)cosomegat` and `x_(2) = l - A_(2)cosomegat` length `= x_(2) - x_(1)` `= l - (A_(1) + A_(2))cosomegat` 
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