question_answer

Time period of a simple pendulum of length (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 oscillations in both the cases is small. Then \[\frac{{{T}_{1}}}{{{T}_{2}}}\] is

A)  \[\frac{1}{\sqrt{3}}\]                                    

B)  \[1\]

C)  \[\sqrt{\frac{4}{3}}\]                    

D)         \[\sqrt{\frac{3}{2}}\]

Correct Answer: D

Solution :

Time period of simple pendulum is given by                 \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] and time period of uniform rod in given position is given by                 \[{{\text{T}}_{\text{2}}}\text{=}\sqrt{\frac{\text{inertia}\,\,\text{factor}}{\text{spring}\,\,\text{factor}}}\] Here, inertia factor = moment of inertia of rod                                           at one end                                       \[=\frac{m{{l}^{2}}}{12}+\frac{m{{l}^{2}}}{4}\]                                       \[=\frac{m{{l}^{2}}}{3}\] Spring factor = restoring torque per unit                              angular displacement                                 \[=mg\times \frac{l}{2}\frac{\sin \theta }{\theta }\]                                 \[=mg\frac{l}{2}\]                            (if\[\theta \]is small) \[\therefore \]  \[{{T}_{2}}=2\pi \sqrt{\frac{m{{l}^{2}}/3}{mgl/2}}=2\pi \sqrt{\frac{2}{3}\frac{l}{g}}\] Hence,  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{3}{2}}\]