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Question #43854

Difficulty: 8/10 | 🟢 Free

The Question

A uniform square plate of mass MM and side length aa is suspended from a frictionless pivot at one of its corners. The plate lies in a vertical plane and is free to oscillate under gravity. It is displaced by a small angle θ0\theta_0 (in radians) from its stable equilibrium position and released from rest. The gravitational restoring torque can be modeled as an effective torsional spring with a rotational spring constant keffk_{eff}. Assuming the acceleration due to gravity is gg, what is the value of keffk_{eff} and the maximum angular velocity ωmax\omega_{max} of the plate?

Options

  • A.

    keff=Mga2k_{eff} = \frac{Mga}{\sqrt{2}}, ωmax=θ03g22a\omega_{max} = \theta_0 \sqrt{\frac{3g}{2\sqrt{2}a}}

  • B.

    keff=Mga2k_{eff} = \frac{Mga}{\sqrt{2}}, ωmax=θ032ga\omega_{max} = \theta_0 \sqrt{\frac{3\sqrt{2}g}{a}}

  • C.

    keff=Mgak_{eff} = Mga, ωmax=θ03g2a\omega_{max} = \theta_0 \sqrt{\frac{3g}{2a}}

  • D.

    keff=Mga2k_{eff} = \frac{Mga}{2}, ωmax=θ03g4a\omega_{max} = \theta_0 \sqrt{\frac{3g}{4a}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Physical Pendulum Inertia|Effective Spring Constant|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to map the gravitational potential energy of a physical pendulum to an equivalent elastic potential energy using an effective rotational spring constant (keffk_{eff}).

The primary trap lies in the calculation of the moment of inertia.

Students frequently forget to apply the Parallel Axis Theorem, incorrectly using the center-of-mass moment of inertia (Icm=Ma2/6I_{cm} = Ma^2/6) instead of the moment of inertia about the pivot (I=2Ma2/3I = 2Ma^2/3).

Additionally, students may misidentify the distance from the pivot to the center of mass, confusing the side length aa or half-side a/2a/2 with the correct diagonal distance a/2a/\sqrt{2}, leading to an incorrect keffk_{eff}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterOscillations - Class 11 Part 2
ConceptPhysical Pendulum Inertia
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree