question_answer

Time period of a simple pendulum of length I 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 given position is given by                 \[{{T}_{2}}=2\pi \sqrt{\frac{inertia\text{ }factor}{spring\text{ }factor}}\] 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                 \[=mg\times \frac{1}{2}\frac{\sin \theta }{\theta }\]                 \[=mg\frac{1}{2}\] (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, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{3}{2}}\]