Ncert Exercises & Solutions

Q.39 Two pendulums with identical bobs and lengths are suspended from 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.

(a) Describe the motion of two bobs.

(b) Draw a graph showing variation in energy of either pendulum with time, for $0 \leq t \leq 2 T$ . where T is the period of each pendulum.

Hint: Both bobs will perform SHM.

Step 1: Interpret about motion.

(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 $E_{1}$ = $E$ . Energy of A, $E_{2}$ = 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 velocity of B. Therefore, $E_{1}$ = 0 and $E_{2}$ = E. At = 2 T/4. B reaches its extreme right when KE of A is converted into PE = $E_{2}$ = E. Energy of $E_{1}$ = 0

At t = 3 T/ 4. A reaches its mean when its PE is converted into KE = $E_{2}$ = E. It collides elastically with B and transfers whole of its energy to B.

Thus, $E_{2} = 0 and E_{1} = E$ . The entire process is repeated.

Step 2: Find energies of different bobs and draw the graph.

(b) The values of energies of B and A at different time intervals are tabulated here. The plot of energy with time $0 \leq t \leq 2 \leq T$ is shown separately for B and A in the figure below.

Time (t) Energy of A $(E_{1})$ Energy of B $(E_{2})$

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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