question_answer

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 oscillations in both the cases is small. Then\[{{T}_{1}}/{{T}_{2}}\]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 give position is given by \[{{T}_{2}}=2\pi \sqrt{\frac{\text{inertia}\,\text{factor}}{\text{spring}\,\text{factor}}}\] 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,                  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{3}{2}}\]