Question

Pendulum $A$ is a physical pendulum made from a thin, rigid and uniform rod whose length is $d$. One end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. Pendulum $B$ is a simple pendulum whose length is also $d$. Obtain the ratio $\frac{T_A}{T_B}$ of their periods for small angle oscillations.

• A. $1:2$
• B. $2:1$
• C. $1:1$
• D. $1:3$

Answer 1

The correct option is C $1:1$

For physical pendulum time period , $T_A=2\pi \sqrt{\frac{I}{mgl}}$ where $I=$ moment of inertia about point O at the end of rod, i.e $I={\int }_0^d{y}^2\left(m/d\right)dy=\frac{m{d}^2}{3}$ and $l=d/2$ distance from O to COM.
thus, $T_A=2\pi \sqrt{\frac{m{d}^2/3}{mg\left(d/2\right)}}=2\pi \sqrt{\frac{2d}{3g}}$
For simple pendulum time period . $T_B=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{d}{g}}$
$\frac{T_A}{T_B}=\sqrt{\left(2d/3g\right)\left(g/d\right)}=0.82\sim 1$