Question

Two pendulums of lengths 100 cm and 121 cm start vibrating. At some instant the two are at the mean position in the same phase. After how many vibrations of the longer pendulum will the two be in the sam phase at the mean position again?

• A. 10
• B. 11
• C. 20
• D. 21

Answer 1

Answer: (A)

Let the two pendulum are in same phase, after n vibrations of
the longer pendulum. In this time the shorter pendulum will
complete ( n + 1) vibrations.
$\therefore n\times 2\pi \sqrt{\frac{l_2}{g}}=(n+1)2\pi \sqrt{\frac{l_1}{g}}$
$n\times 2\pi \sqrt{\frac{121}{g}}=(n+1)2\pi \sqrt{\frac{100}{g}}$
$11n=10(n+1)$
$n=10$