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QJEEM19Ph4756: JEE Main 2019 Physics - Class 11 JEEM19Ph4756 Solution: Two light identical springs of spring constant k are attached…

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length ℓ\ell and mass m. The rod is pivoted at its centre '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 :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Simple Harmonic Motion Question 109 English

Multiple Choice Options

  • A 12π3km{1 \over {2\pi }}\sqrt {{{3k} \over m}}
  • B 12π6km{1 \over {2\pi }}\sqrt {{{6k} \over m}}
  • C 12πkm{1 \over {2\pi }}\sqrt {{k \over m}}
  • D 12π2km{1 \over {2\pi }}\sqrt {{{2k} \over m}}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 12th January Morning Slot Physics - Simple Harmonic Motion Question 109 English Explanation
τ=−2kxℓ2cos⁡θ\tau = - 2kx{\ell \over 2}\cos \theta

⇒\Rightarrow     τ=(Kℓ22)θ=−Cθ\tau = \left( {{{K{\ell ^2}} \over 2}} \right)\theta = - C\theta

⇒\Rightarrow     f=12πC1=12πKℓ22Mℓ212f = {1 \over {2\pi }}\sqrt {{C \over 1}} = {1 \over {2\pi }}\sqrt {{{{{K{\ell ^2}} \over 2}} \over {{{M{\ell ^2}} \over {12}}}}}

⇒\Rightarrow     f=12π6KMf = {1 \over {2\pi }}\sqrt {{{6K} \over M}}

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