Q. Two pendulums of length 212 cm and 100 cm start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase in the mean position?
A) 10
B) 11
C) 20
D) 21
Explanation
(b) $T_{1} = 2\pi\sqrt{\frac{121}{g}}$ and $T_{2} = 2\pi\sqrt{\frac{100}{g}}$
So$T_{1} > T_{2}$. Let the shorter pendulum makes $n$ vibrations, then the longer pendulum will make one less than $n$ vibrations to come in phase again
So $nT_{2} = (n - 1)T_{1}$
Or $n \times 2\pi\sqrt{\frac{100}{g}} = (n - 1) \times 2\pi\sqrt{\frac{121}{g}}$
Or $10n = (n - 1)11$ or $n = 11$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.