Question

Time period of a simple pendulum of length $l$ is $T_{1}$, and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_{2}$. Amplitude of oscillation in both the cases is small. Then $T_{1}/T_{2}$ is

• A. $1/ \sqrt{3}$
• B. $1$
• C. $\sqrt{4 /3}$
• D. $\sqrt{3/ 2}$

Answer 1

The Correct Option is D

Time period of a simple pendulum of length is given by $T_{1}=2\pi\sqrt{l /g}$ Time period of a uniform rod of length $l$ pivoted at one end and oscillating in vertical plane is given by $T_{2}=2\pi \sqrt{\frac{\text{inertia factor}}{\text{spring factor}}}$ Here, inertia factor = moment of inertia of rod at one end $=\frac{ml^{2}}{12}+\frac{ml^{2}}{4}=\frac{ml^{2}}{3}$ Spring factor = restoring torque per unit angular displacement $=\frac{mg\times\frac{l}{2}sin\,\theta}{\theta}=mg \frac{l}{2}$ (if $\theta$ is small) $\therefore T_{2}=2\pi\sqrt{\frac{ml^{2} /3}{mgl /2}}=2\pi \sqrt{\frac{2I}{3g}}$ $\therefore \frac{T_{1}}{T_{2}}=\sqrt{\frac{3}{2}}$