Q. Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. the minimum time after which the two threads of the pendulum will be one behind the other is?
A) $T/4$
B) $T/3$
C) $4\ T/3$
D) $4\ T$
Explanation
(c) $\frac{T_{1}}{T} = \sqrt{\frac{l_{1}}{l}} = \sqrt{\frac{16}{1}} = 4$
Or $t_{1} = 4T$
Let after time$t$, the pendulum be in the same phase. It will be so then
$\frac{t}{T_{1}} = \frac{t}{T} - 1 = \frac{t - T}{T}$
or $\frac{t}{4T} = \frac{t - T}{T}$
or $t = 4t - 4t$
or $3t = 4T$
or $t = 4T/3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.