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Question #83319

Difficulty: 5/10 | 🟢 Free

The Question

A mechanical oscillator consists of a block of mass $m = 1.0 \text{ kg}$ attached to a viscoelastic spring of stiffness $k = 100 \text{ N/m}$ . Due to elastic hysteresis, the spring dissipates energy. For an oscillation cycle with amplitude $x_0$ , the area of the hysteresis loop in the force-extension graph is given by $A = \pi x_0^2 \text{ J}$ . This mechanical system is modeled as an equivalent series $LCR$ circuit, where the mass corresponds to the inductance $L$ , the spring stiffness corresponds to the inverse of capacitance $1/C$ , and the hysteresis loss is modeled by an equivalent constant resistance $R$ . Assuming the damping is small, which of the following statements regarding the system is INCORRECT?

Options

A.
The equivalent electrical resistance of the circuit is $R = 0.1 \text{ \Omega}$ .
B.
The Quality factor ( $Q$ -factor) of the equivalent circuit is $100$ .
C.
The amplitude of the mechanical oscillations decreases by a factor of $e^{-1}$ after a time interval of $10 \text{ s}$ .
D.
The average rate of energy dissipation per unit time is $5 x_0^2 \text{ W}$ , where $x_0$ is the amplitude of that cycle.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Elastic Hysteresis Loop|LC Circuit Oscillations|Integration

Why This Question Is Tricky

Examiner Levers

Negative Thinking

Uses "NOT", "EXCEPT", or "INCORRECT" to flip the logic gate.

Difficulty Boost : +2

Common Mistakes

This question tests the mapping of mechanical hysteresis (energy loss per cycle) to electrical resistance (energy loss per unit time) in an equivalent $LCR$ circuit.

The primary trap lies in distinguishing between the exponential decay constant of energy versus amplitude.

Students often incorrectly apply the energy decay time constant ( $\tau_E = L/R = 10 \text{ s}$ ) to the amplitude decay, forgetting that energy is proportional to the square of the amplitude, which introduces a factor of 2 in the exponent ( $\tau_A = 2L/R = 20 \text{ s}$ ).

By using 'INCORRECT' as the lever, the student is forced to rigorously verify all four interconnected parameters.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Elastic Hysteresis Loop
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free