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JEE Advanced 2023 Paper 1 · Physics · Simple Harmonic Motion · medium

A simple pendulum of length L has period T0=2π√(L/g) when oscillating through small angles in a fixed laboratory. List-I describes four situations of…

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A simple pendulum of length LL has period T0=2πL/gT_0=2\pi\sqrt{L/g} when oscillating through small angles in a fixed laboratory. List-I describes four situations of the same pendulum and List-II gives the ratio T/T0T/T_0 of the new period to T0T_0. Match List-I with List-II.

List-I
(P) Pendulum in a lift accelerating upward at g/3g/3
(Q) Pendulum in a vehicle accelerating horizontally at 3 g\sqrt3\,g
(R) Pendulum whose string, when vertical, is caught at its midpoint by a fixed peg (small oscillations on both sides)
(S) Pendulum in a lift accelerating downward at g/2g/2

List-II
(1) 2\sqrt2
(2) 12\dfrac{1}{\sqrt2}
(3) 32\dfrac{\sqrt3}{2}
(4) 2+24\dfrac{2+\sqrt2}{4}
(5) 22
  1. A. P→3, Q→2, R→4, S→1
  2. B. P→3, Q→2, R→1, S→4
  3. C. P→5, Q→2, R→4, S→1
  4. D. P→3, Q→4, R→2, S→1
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Answer: (A) P→3, Q→2, R→4, S→1

The period is 2πL/geff2\pi\sqrt{L/g_{eff}}, so T/T0=g/geffT/T_0=\sqrt{g/g_{eff}}. (P) geff=g+g/3=43gg_{eff}=g+g/3=\tfrac43g, so T/T0=3/4=32T/T_0=\sqrt{3/4}=\dfrac{\sqrt3}{2}, which is (3). (Q) geff=g2+3g2=2gg_{eff}=\sqrt{g^2+3g^2}=2g, so T/T0=12T/T_0=\dfrac1{\sqrt2}, which is (2). (R) The complete oscillation consists of half a swing with length LL and half a swing with length L/2L/2, so T=12T0+12⋅T02T=\tfrac12T_0+\tfrac12\cdot\dfrac{T_0}{\sqrt2}, so TT0=12(1+12)=2+24\dfrac{T}{T_0}=\dfrac12\left(1+\dfrac1{\sqrt2}\right)=\dfrac{2+\sqrt2}{4}, which is (4). (S) geff=g−g/2=g/2g_{eff}=g-g/2=g/2, so T/T0=2T/T_0=\sqrt2, which is (1). Item (5) is unused. Thus P→3, Q→2, R→4, S→1.
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