InterviewSolution
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InterviewSolution
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A block with mass `M` attached to a horizontal spring with force constant `k` is moving with simple harmonic motion having amplitude `A_(1)`. At the instant when the block passes through its equilibrium position a lump of putty with mass `m` is dropped vertically on the block from a very small height and sticks to it. (a) Find the new amplitude and period. (b) Repeat part (a) for the case in which the putty is dropped on the block when it is at one end of its path. |
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Answer» (a) Before the lump of putty is dropped the total mechinical energy fo the blcok and string is `E_(1) = (1)/(2) kA_(1)^(2)`. Since, the block is at the equilibrium position `U = 0`, and the energy is purely kinetic. Let `v_(1)` be the speed of the block at the equilibrium position, we have `E_(1) = (1)/(2) Mv_(1)^(2) = (1)/(2) kA_(1)^(2)` `v_(1) =sqrt((k)/(M)) A_(1)` During the process momentum of the system in horizontal direction is conserved. Let `v_(2)`be the speed of the combined mass, then `(M +m) v_(2) = Mv_(1)` `v_(2) =(M)/(M+m) v_(1)` Now, let `A_(2)` be the amplitude afterwards. Then, `E_(2) = (1)/(2) kA_(2)^(2) = (1)/(2) (M +m) v_(2)^(2)` Substiting the proper vales, we have `A_(2) = A_(1) sqrt((M)/(M+m))` Note: `E_(2) lt E_(1)`, as some energy is lost into heating up the block and putty. Further, `T_(2)=2pi sqrt((M+m)/(k))` (b) When the putty drops on the block, the block is intaneously at rest. All the mechanical enegry is strored in the spring as potential enegry. Again the momentum in horizontal direction is conserved during the process. But now it is zero just before and after putty is dropped. So, in this case, adding the extra mass of the putty has no effect on the mechanical energy, i.e., `E_(2) = E_(1) = (1)/(2) kA_(1)^(2)` and the amplitudes is still `A_(1)`. Thus, `A_(2) = A_(1)` and `T_(2) = 2pi sqrt((M+m)/(k))` |
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