Question

Two pendulums of length $121cm$ and $100cm$ start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

• A. 8
• B. 11
• C. 9
• D. 10

Answer 1

Answer: (B)

$(n)T_{\ell }=(n+1)T_s$
$(n)2\pi \sqrt{\frac{1.21}{g}}=(n+1)2\pi \sqrt{\frac{1}{g}}$
$(n)(1.1)=(n+1)$
$0.1(n)=1$
$n=10$
No. of oscillation of smaller one
$=n+1$
$=10+1$
$=11$