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प्रश्न
Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (Figure). One of the bobs is released after being displaced by 10° so that it collides elastically head-on with the other bob.
- Describe the motion of two bobs.
- Draw a graph showing variation in energy of either pendulum with time, for 0 ≤ t ≤ 2T, where T is the period of each pendulum.
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उत्तर
a. Consider the adjacent diagram in which the bob B is displaced through an angle θ and released.
At t = 0, suppose bob B is displaced by θ = 10° to the right. It is given potential energy E1 = E. Energy of A, E2 = 0
When B is released, it strikes A at t = T/4. In the head-on elastic collision between B and A comes to rest and A gets the velocity of B. Therefore. E1 = 0 and E2 = E. At t = 27T/4, B reaches its extreme right position when KE of A is converted into PE = E2 = E. Energy of B, E1 = 0.
At t = 3T/4, A reaches its mean position. when its PE is converted into KE = E2 = E. It collides elastically with B and transfers the whole of its energy to B. Thus, E2 = 0 and E1 = E. The entire process is repeated.
b. The values of energies of B and A at different time intervals are tabulated here. The plot of energy with time 0 ≤ t ≤ 2T is shown separately for 8 and A in the figure below.
| Time (t) | Energy of A (E1) |
Energy of B (E2) |
| 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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