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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Physics
Chapter
Oscillations

JEE Advanced 2009 · Paper 2 · Physics · Oscillations

A uniform rod of length L and mass M is pivoted at the centre. Its...

A uniform rod of length LL and mass MM is pivoted at the centre. Its two ends are attached to two springs of equal spring constants kk. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ\theta in one direction and released. The frequency of oscillation is A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring

Options

  1. A12π2kM\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}
  2. B12πkM\frac{1}{2 \pi} \sqrt{\frac{k}{M}}
  3. C12π6kM\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}
  4. D12π24kM\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}

Correct answer

C. 12π6kM\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}

Step-by-step solution

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring
 Restoring torque =−(2kx)⋅L2α=−kL(L/2θ)I=−[kL2/2ML2/12]⋅θ\begin{aligned}& \text{ Restoring torque }=-(2 k x) \cdot \frac{L}{2}\\ & \qquad \alpha=-\frac{k L(L / 2 \theta)}{I}=-\left[\frac{k L^2 / 2}{M L^2 / 12}\right] \cdot \theta \end{aligned}=−(6kM)θ∴f=12π∣αθ∣=12π6kM\begin{aligned}& =-\left(\frac{6 k}{M}\right) \theta\\ \therefore \quad f & =\frac{1}{2 \pi} \sqrt{\left|\frac{\alpha}{\theta}\right|}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}} \end{aligned}
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