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Question #42035

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ (in $\text{kg}$ ) is constrained to move along the $x$ -axis. It is subjected to a restoring force $F = -kx - \alpha x^3$ , where $k$ (in $\text{N m}^{-1}$ ) and $\alpha$ (in $\text{N m}^{-3}$ ) are positive constants. If the particle is released from rest at position $x = A$ (in $\text{m}$ ), what is its speed $v$ (in $\text{m s}^{-1}$ ) when it passes through the origin ( $x = 0$ )?

Options

A.
$\sqrt{\frac{k}{m} A^2 + \frac{\alpha}{2m} A^4}$
B.
$\sqrt{\frac{k}{m} A^2 + \frac{\alpha}{m} A^4}$
C.
$A \sqrt{\frac{k + \alpha A^2}{m}}$
D.
$A \sqrt{\frac{k}{m}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

The Velocity-Position Identity|SHM Definition via Differential Equation|Binomial Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize when standard Simple Harmonic Motion (SHM) kinematics fail.

The presence of the cubic term $\alpha x^3$ creates an anharmonic (Duffing-type) potential.

A common trap is reflexively applying the SHM identity $v = \omega\sqrt{A^2-x^2}$ by either ignoring the non-linear term or incorrectly defining an 'effective' spring constant $k_{eff} = k + \alpha A^2$ .

The correct approach requires abandoning memorized formulas and returning to first principles by integrating the work-energy theorem ( $F = mv \frac{dv}{dx}$ ) to find the exact change in kinetic energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	The Velocity-Position Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free