Physics - SHM Question with Solution | TestHub

Let the time elapsed be $t$.

For the two pendulums to be in phase again, the time $t$ must be an integer multiple of both their time periods.

$t=n_1{T}_1=n_2{T}_2$

Given $T_1=T$ and $T_2=\frac{4T}{3}$.

$n_{1}T=n_2\left(\frac{4T}{3}\right)$

$\frac{n_1}{n_2}=\frac{4}{3}$

For the first time they are in phase again, the difference in the number of oscillations is $1$.

Thus, $n_1=4$ and $n_2=3$.

The time elapsed is $t=4T$.

Alternatively, the time elapsed is the least common multiple of the two time periods.

$t=\text{LCM}\left(T,\frac{4T}{3}\right)=4T$