PrepRung
Sign inSign up free

JEE Advanced 2018 Paper 1 · Physics · Simple Harmonic Motion · medium

A uniform thin rod of mass 1.0 kg and length 1.0 m hangs vertically from a smooth horizontal hinge at its upper end. A light horizontal spring of…

Numerical answer

A uniform thin rod of mass 1.0 kg1.0\ \mathrm{kg} and length 1.0 m1.0\ \mathrm{m} hangs vertically from a smooth horizontal hinge at its upper end. A light horizontal spring of force constant 10 N m−110\ \mathrm{N\,m^{-1}} is attached to the lower end of the rod, its other end being fixed to a wall, and the spring is unstretched when the rod is vertical. The rod is displaced through a small angle and released. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. Find the period of small oscillations in seconds. Give the answer correct to two decimal places.
Show answer and solution

Answer: 0.94

For a small angle θ\theta about the hinge, two restoring torques act: gravity gives mgL2θ=5θmg\dfrac L2\theta=5\theta (in N m) and the spring, stretched by LθL\theta with force kLθkL\theta acting at distance LL, gives kL2θ=10θkL^2\theta=10\theta. The moment of inertia about the hinge is I=mL23=13 kg m2I=\dfrac{mL^2}{3}=\dfrac13\ \mathrm{kg\,m^2}. The torque equation Iθ¨=−(5+10)θI\ddot\theta=-(5+10)\theta is simple harmonic with ω2=15/(1/3)=45 s−2\omega^2=15/(1/3)=45\ \mathrm{s^{-2}}. Hence T=2π/45=0.9366 s≈0.94 sT=2\pi/\sqrt{45}=0.9366\ \mathrm{s}\approx0.94\ \mathrm{s}. Neglecting either the gravitational torque or the spring torque would give a wrong period.
Report a mistake

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake