question_answer

A uniform square lamina of side 2a is hung up by one comer and it oscillates in its own plane which is vertical. The length of the equivalent simple pendulum is

A)  \[\ell =\frac{a}{\sqrt{2}}\]                          

B)  \[\ell =\frac{\sqrt{2}}{3}a\]

C)  \[\ell =\frac{4\sqrt{2}}{3}a\]                     

D)  \[\ell =4\sqrt{2}\,s\]

Correct Answer: C

Solution :

When the lamina ABCD is at rest, its centre of gravity G must lie vertically below the comer A by which it is hung.                 By geometry, \[AG=\sqrt{2}a\] Radius of gyration about this axis is given by \[{{K}^{2}}=\frac{2}{3}{{a}^{2}}\left( {{I}_{G}}=\frac{2}{3}m{{a}^{2}} \right)\] Then length of equivalent simple pendulum is: \[\ell =\sqrt{2}a+\frac{\frac{2}{3}{{a}^{2}}}{\sqrt{2}\,a}=\frac{4\sqrt{2}}{3}a\]