JEE Main & Advanced
Physics
Simple Harmonic Motion
Question Bank
Critical Thinking
question_answer
A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
A) \[\frac{5}{4}R\]
B) \[\frac{2}{3}R\]
C) \[\frac{3}{4}R\]
D) \[\frac{3}{2}R\]
Correct Answer:
D
Solution :
Time period of a physical pendulum \[T=2\pi \sqrt{\frac{{{I}_{0}}}{mgd}}=2\pi \sqrt{\frac{\left( \frac{1}{2}m{{R}^{2}}+m{{R}^{2}} \right)}{mgR}}\] \[=2\pi \sqrt{\frac{3R}{2g}}\] ?..(i) \[{{T}_{simple\ pendulum}}=2\pi \sqrt{\frac{l}{g}}\] ?..(ii) Equating (i) and (ii), \[l=\frac{3}{2}R.\]