Q.28 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 28 Solution

Topic: Oscillations & SHM · Official NTA Booklet Code 50

Question

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k N m⁻¹. At a given instant, the extension of the spring is x meter and the speed of the particle is v m s⁻¹. On the x-v plane, if the graph of v as a function of x is a circle, then the correct option is:

✓

Verified Against NTA Final Answer Key

Option (2) : k = m

📚 Oscillations & SHM Medium

📋 Given Parameters & Problem Setup

Oscillator 1D spring-mass system with mass m kg and spring constant k N/m

Instantaneous State Displacement x meters, velocity v m/s

Phase Plane v as a function of x on the x - v plane

Geometric Condition Graph of v vs x is a perfect circle

Target Relation Relation between spring constant k and mass m

⚡

The 30-Second Exam Hall Shortcut: ⚡ Phase Trajectory Aspect Ratio Shortcut

⏱️ Instant conceptual recognition.

A phase space ellipse becomes a circle when omega = 1 rad/s!

1 Velocity in SHM: v^2 = omega^2 * (A^2 - x^2) ==> v^2 + omega^2 * x^2 = omega^2 * A^2.

2 Divide by omega^2 * A^2: (x / A)^2 + (v / (omega*A))^2 = 1.

3 This is an ellipse with semi-axes a = A and b = omega * A.

4 For an ellipse on the x-v plane to be a circle, semi-axes must be equal: A = omega * A ==> omega = 1.

5 Since omega = sqrt(k / m): sqrt(k / m) = 1 ==> k / m = 1 ==> k = m. Solved in 10 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Conservation of mechanical energy: (1/2)*m*v^2 + (1/2)*k*x^2 = E. Normalizing gives x^2 / (2E/k) + v^2 / (2E/m) = 1. A circle requires equal denominators: 2E/k = 2E/m ==> k = m.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulate Conservation of Mechanical Energy in SHM

For an undamped spring-mass oscillator of mass m and spring constant k, total mechanical energy E is conserved:

E = \frac{1}{2} m v^2 + \frac{1}{2} k x^2

• x is the extension/compression from equilibrium in meters.
• v is the instantaneous velocity in m/s.

Step 2

Express the Energy Equation in Standard Ellipse Form

Divide both sides of the energy equation by total energy E:

\frac{x^2}{\left( \frac{2E}{k} \right)} + \frac{v^2}{\left( \frac{2E}{m} \right)} = 1

• This represents an ellipse on the x - v plane with horizontal semi-axis a = \sqrt{2E/k} and vertical semi-axis b = \sqrt{2E/m}.

Step 3

Enforce the Geometric Condition for a Circle

An ellipse becomes a circle if and only if both semi-axes are equal in magnitude (a = b):

\sqrt{\frac{2E}{k}} = \sqrt{\frac{2E}{m}} \implies \frac{2E}{k} = \frac{2E}{m} \implies k = m

• Equivalently, angular frequency omega = sqrt(k/m) = 1 rad/s.
• Matches Option (2) exactly.

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Reciprocal relation.

⚠️

Student Fallacy (Why candidates pick this trap): Confuses omega^2 = k/m with omega = 1/(k*m).

Note: Circle requires k = m, not k = 1/m.

2 Option (2)

✓ 100% Correct

Verdict & Scientific Reason:

100% mathematically exact; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Correctly equates ellipse semi-axes: 2E/k = 2E/m ==> k = m.

Note: Exact physics.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Quadratic relation.

⚠️

Student Fallacy (Why candidates pick this trap): Squares mass unnecessarily.

Note: k/m = 1 implies linear equality k = m.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Square root relation.

⚠️

Student Fallacy (Why candidates pick this trap): Confuses omega with k.

Note: omega = sqrt(k/m) = 1 ==> k = m.

⚠️

NTA Trap Alert: Dimensional Analysis Misconception

Students often object: "k and m have different units (N/m vs kg), so how can k = m?"

🛡️ How to avoid the -1 mark: The question explicitly specifies "on the x - v plane, if the graph is a circle" — this is a numerical scale equality of the coordinate axes, which requires the numerical value of k to equal m (omega = 1 rad/s).

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Oscillations Standard Benchmark

Core Scientific Principle Tested:

"Total energy: E = (1/2)*m*v^2 + (1/2)*k*x^2. Phase space trajectory is an ellipse x^2/(2E/k) + v^2/(2E/m) = 1. Circle condition: k = m (omega = 1)."

Syllabus Focus: Energy in Simple Harmonic Motion and Phase Trajectories

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 28 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Phase Trajectory Aspect Ratio Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulate Conservation of Mechanical Energy in SHM

Express the Energy Equation in Standard Ellipse Form

Enforce the Geometric Condition for a Circle

360° Option Elimination Engine Key Feature

NTA Trap Alert: Dimensional Analysis Misconception

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Oscillations

Spring-Mass System with Slowly Leaking Mass (Variable Mass SHM)

Ratio of Amplitudes for Identical Masses with Equal Maximum Speeds

Amplitude and Time Period of an SHM Equation

Time Period Variation with Mass and Length of a Simple Pendulum

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