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Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position, the two bobs are in constact, . One of the bobs is released after being displaced by `10^(@)` so that it collides elastically head - on with the other bob. (a) Describe the motion of two bobs. (b) Draw a graph showing variation in energy of either pendulum with time,for `0letle2T` , where `T` is the period of each pendulum. |
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Answer» As two pendulum have identical bobs and lengths, they will execute simple harmonic motions of same time period, say `T`. At `t=0`, suppose bob `A` is displaced by `10^(@)` to the left. It is given potential energy `E_(1)=E` . Energy of `b,e_(2)=0`. When `A` is released, it strikes `B` at `t=T//4` . In the head on elastic collision between `A` and `B`, `A` comes to rest and `B` gets velcoity of `A`. Therefore, `E_(1)=0` and `E_(2)=E`. At `t=2T//4`, `B` reahes its extreme right position when `K.E.` of `B` is converted into `P.E.=E_(2)=E`. Energy of `A,E_(1)=0.`At `t=3T//4,` `B` reaches its mean position, when its P.E. is converted into `KE.=E_(2)=E`. It collides elastically with `A` and transfers whole of its energy to `A` . Thus, `E_(2)=0` and `E_(1)=E`. The entire process is repeated. The value of energies of `A` and `B` at different time intervals are tabulated here. The plot of energy with time `0letle2T` is shown separately for `A` and `B` in figure. `|{:(time(t),"Energy of" A,"Energy of" AB,),(,E_(1),E_(2),),(0,E,0,),(T//4,0,E,),(2T//4,0,E,),(3T//4,E,0,),(4T//4,E,0,),(5T//4,0,E,),(6T//4,0,E,),(7T//4,E,0,),(8T//4,E,0,):}|` |
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