Question

Two simple pendulums of lengths $100$ cm and $196$ cm are in phase at the mean position at a certain time. If $T$ is the time period of shorter pendulum, the minimum time after which they will be again in phase

• A. $2.5T$
• B. $3.5T$
• C. $5T$
• D. $7T$

Answer 1

Answer: (B)

$3.5T$

$T=2\pi \sqrt{\frac{l}{g}}$
$T_2=2\pi \sqrt{\frac{l_1}{g}}$
$T_1=2\pi \sqrt{\frac{100}{g}}$
$T_2=2\pi \sqrt{\frac{l}{g}}$
$T_2=2\pi \sqrt{\frac{196}{g}}$
$\frac{T_1}{T_2}=\frac{5}{7}$
$\therefore ({\omega }_1-{\omega }_2)t=2\pi$
$(\frac{2\pi }{T_1}-\frac{2\pi }{T_2})t=2\pi$
$t=\frac{7}{2}$
$t=3.5sec$