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JEE Main 2022 · Physics · Simple Harmonic Motion · medium

A uniform thin rod of length L is suspended from a horizontal axis perpendicular to the rod and passing through a point at distance L/4 from one end.…

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A uniform thin rod of length LL is suspended from a horizontal axis perpendicular to the rod and passing through a point at distance L4\dfrac L4 from one end. For small oscillations in a vertical plane, the time period is
  1. A. 2πLg2\pi\sqrt{\dfrac{L}{g}}
  2. B. 2π2L3g2\pi\sqrt{\dfrac{2L}{3g}}
  3. C. 2π7L12g2\pi\sqrt{\dfrac{7L}{12g}}
  4. D. 2π5L12g2\pi\sqrt{\dfrac{5L}{12g}}
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Answer: (C) 2π7L12g2\pi\sqrt{\dfrac{7L}{12g}}

The rod is a physical pendulum with pivot at distance d=L2−L4=L4d=\dfrac L2-\dfrac L4=\dfrac L4 from the centre of mass. By the parallel-axis theorem I=mL212+m(L4)2=7mL248I=\dfrac{mL^{2}}{12}+m\left(\dfrac L4\right)^{2}=\dfrac{7mL^{2}}{48}. For small angles T=2πImgd=2π7L2/48gL/4=2π7L12gT=2\pi\sqrt{\dfrac{I}{mgd}}=2\pi\sqrt{\dfrac{7L^{2}/48}{gL/4}}=2\pi\sqrt{\dfrac{7L}{12g}}. Using I=mL23I=\frac{mL^2}{3} (end pivot) would give 2π2L/3g2\pi\sqrt{2L/3g}.
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