$\omega=\sqrt{\dfrac{g}{l}}$
$\Rightarrow T=\dfrac{2\pi}{\omega}=2\pi.\sqrt{\dfrac{l}{g}}$
Khi $l'=\dfrac{l}{2,25}=\dfrac{4l}{9}$:
$T'=2\pi.\sqrt{\dfrac{4l}{9g}}$
$=2\pi.\dfrac{2}{3}\sqrt{\dfrac{l}{g}}$
$=\dfrac{2}{3}T$
Vậy chu kì giảm $\dfrac{2}{3}$ lần.