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JEE Advanced 2023 Paper 2 · Physics · Simple Harmonic Motion · medium

JEE Advanced 2023 Paper 2 Physics: Paragraph 2

A uniform thin rod of length L=1.20 mL=1.20\ \mathrm{m} and mass mm can oscillate in a vertical plane as a physical pendulum about a smooth horizontal axis perpendicular to the rod and passing through a point at a distance dd from its centre. For small oscillations take g=9.8 m s−2g=9.8\ \mathrm{m\,s^{-2}}. The moment of inertia of the rod about an axis through its centre perpendicular to its length is mL212\dfrac{mL^2}{12}.

Question 1

Numerical answer

The rod is pivoted at one of its ends (so d=0.60 md=0.60\ \mathrm{m}). Find the time period of small oscillations, in seconds. Give the answer correct to two decimal places.
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Answer: 1.80

By the parallel-axis theorem, I=mL212+md2I=\dfrac{mL^2}{12}+md^2, and the restoring torque for small angles is mgd θmgd\,\theta. Hence T=2πImgd=2πL2/12+d2gdT=2\pi\sqrt{\dfrac{I}{mgd}}=2\pi\sqrt{\dfrac{L^2/12+d^2}{gd}}. For L=1.20 mL=1.20\ \mathrm{m}, d=0.60 md=0.60\ \mathrm{m}: L2/12=0.12L^2/12=0.12, so numerator =0.12+0.36=0.48 m2=0.12+0.36=0.48\ \mathrm{m^2}, denominator =9.8×0.60=5.88=9.8\times0.60=5.88. T=2π0.48/5.88=2π0.08163=2π(0.28571)=1.795 s≈1.80 sT=2\pi\sqrt{0.48/5.88}=2\pi\sqrt{0.08163}=2\pi(0.28571)=1.795\ \mathrm{s}\approx1.80\ \mathrm{s}. This equals the period of a simple pendulum of length ℓ=L2/12d+d=0.80 m\ell=\dfrac{L^2/12}{d}+d=0.80\ \mathrm{m}, i.e. 2L/32L/3 as expected for a rod pivoted at its end.

Question 2

Numerical answer

The pivot position dd can be chosen freely along the rod. Find the minimum possible time period of small oscillations, in seconds. Give the answer correct to two decimal places.
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Answer: 1.67

The period T=2πℓgT=2\pi\sqrt{\dfrac{\ell}{g}} with equivalent length ℓ(d)=L212d+d\ell(d)=\dfrac{L^2}{12d}+d. Minimising, dℓdd=−L212d2+1=0\dfrac{d\ell}{dd}=-\dfrac{L^2}{12d^2}+1=0 gives d=L12=0.3464 md=\dfrac{L}{\sqrt{12}}=0.3464\ \mathrm{m} (the radius of gyration about the centre). Then ℓmin=2d=L3=0.6928 m\ell_{min}=2d=\dfrac{L}{\sqrt3}=0.6928\ \mathrm{m}. Therefore Tmin=2π0.69289.8=2π0.07070=2π(0.26589)=1.671 s≈1.67 sT_{min}=2\pi\sqrt{\dfrac{0.6928}{9.8}}=2\pi\sqrt{0.07070}=2\pi(0.26589)=1.671\ \mathrm{s}\approx1.67\ \mathrm{s}, less than the period 1.80 s1.80\ \mathrm{s} when the pivot is at the end.
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