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  • Solve this problem Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. Th

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

  • Option 1)

     \frac{1}{2 \pi} \sqrt{\frac{k}{m}}

  • Option 2)

    \frac{1}{2 \pi} \sqrt{\frac{6k}{m}}

  • Option 3)

    \frac{1}{2 \pi} \sqrt{\frac{3k}{m}}

  • Option 4)

    \frac{1}{2 \pi} \sqrt{\frac{2k}{m}}

Answers (1)

best_answer

 

Time Period of Torsional pendulum case -

T=2\pi \sqrt{\frac{I}{K}}

- wherein

I= moment of inertia

K= torsional constant

From Figure 

x=\frac{l}{2}sin\theta =\left ( \frac{l}{2}\times \theta \right )

& \iota =\underset{F_{x}}{\rightarrow}.\underset{S}{\rightarrow}

=-2kx\times \left ( \frac{l}{2} \right )

\iota =-2\times k\times \frac{l}{2}\Theta \times \frac{l}{2}=I\alpha =\frac{ml^{2}}{12}\alpha

\Rightarrow \frac{Ke^{2}}{2}\Theta =\frac{-ml^{2}}{12}\alpha

\alpha =\frac{6k}{m}\Theta =\, \, \, \, so\, \, w=\sqrt{\frac{6k}{m}}

f=\frac{w}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{6k}{m}}

 

 

 


Option 1)

 \frac{1}{2 \pi} \sqrt{\frac{k}{m}}

Option 2)

\frac{1}{2 \pi} \sqrt{\frac{6k}{m}}

Option 3)

\frac{1}{2 \pi} \sqrt{\frac{3k}{m}}

Option 4)

\frac{1}{2 \pi} \sqrt{\frac{2k}{m}}

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