Class NEET 2026 ALL Q #1946
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k~Nm^{-1}$. At a given instant, the extension of the spring is $x$-meter and the speed of the particle is $v~ms^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:
(A) $k=\sqrt{m}$
(B) $k=\frac{1}{m}$
(C) $k=m$
(D) $k=m^{2}$
Correct Answer: C

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Detailed Solution

Step 1: Energy Conservation Principle

For a simple harmonic oscillator, the total mechanical energy E is conserved and is given by the sum of kinetic energy and potential energy: $$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$$

Step 2: Rearranging into Standard Equation

Divide the entire equation by E to get the form of an ellipse: $$\frac{v^2}{2E/m} + \frac{x^2}{2E/k} = 1$$

Step 3: Condition for a Circle

For the graph in the x-v plane to be a circle, the coefficients of x^2 and v^2 must be equal, or more simply, the denominators must be equal: $$\frac{2E}{m} = \frac{2E}{k}$$ This implies m = k.

Final Answer: k=m

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must translate the physical principle of energy conservation into a mathematical geometric constraint.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a sequence of algebraic steps to transform the energy equation into the standard form of a conic section.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. It tests the ability to visualize phase space trajectories, which is a higher-order skill required for competitive physics.