Question

The time periods of oscillation of two simple pendulums are 1 sec, 1.2 sec. Initially both are in same phase of oscillation. The minimum number of oscillations made by longer pendulum when they are again in same phase is:

• A. 5
• B. 6
• C. 10
• D. 12

Answer 1

The correct option is A 5

$t({\omega }_1-{\omega }_2)=2\pi$
$t=\frac{T_1{T}_2}{T_2-T_1}$
$=\frac{1\times 1.2}{1.2-1}$
$t=6$
$\therefore$ no. of oscillations made lag longer pendulum $=\frac{6}{1.2}$
$=5$