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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2019 · 12 January · Morning Shift · Physics · Oscillations

Two light identical springs of spring constant k are attached...

Two light identical springs of spring constant kk are attached horizontally at the two ends of a uniform horizontal rod ABAB of length ll and mass m.m . The rod is pivoted at its center 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod

Options

  1. A12π3km\dfrac{1}{2\pi}\sqrt{\dfrac{3k}{m}}
  2. B12πkm\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}
  3. C12π6km\dfrac{1}{2\pi}\sqrt{\dfrac{6k}{m}}
  4. D12π2km\dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}

Correct answer

C. 12π6km\dfrac{1}{2\pi}\sqrt{\dfrac{6k}{m}}

Step-by-step solution

Torque on the rod about O,O,

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod

τ=kx.l2×2τ = kx . \dfrac{l}{2} \times 2

Iα=k.l2.θ.l2×2I \alpha = k . \dfrac{l}{2} . \theta . \dfrac{l}{2} \times 2

Iα=kl22.θI \alpha = k\dfrac{{l}^{2}}{2} . \theta

∴ω2=kl22Ml212=6kM∴ {\omega}^{2} = \dfrac{k\dfrac{{l}^{2}}{2}}{M\dfrac{{l}^{2}}{12}} = \dfrac{6k}{M}

f=12π6kMf = \dfrac{1}{2\pi}\sqrt{\dfrac{6k}{M}}
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