Q.27 Physics ⭐ 360° Option Elimination Active

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NEET 2021 Physics Question 27 Solution

Topic: Oscillations · Official NTA Booklet Code M1

Question

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

✓

Verified Against NTA Final Answer Key

Option (2) : 2n

📚 Oscillations Easy

📋 Given Parameters & Problem Setup

Frequency of SHM Motion n

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The 30-Second Exam Hall Shortcut: ⚡ Double Frequency Rule: Energy Completes 2 Cycles per 1 SHM Cycle

⏱️ Save ~20 seconds

In one full oscillation, the particle passes through the equilibrium position twice and reaches extreme positions twice. Hence, both potential energy and kinetic energy go through two complete cycles of oscillation for every one cycle of displacement: Frequency of PE $= 2n$ ! Match Option (2)!

1 Displacement:

x(t) = A \sin(2\pi n t)

2 Potential Energy:

U(t) = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(2\pi n t)

3 Use trigonometric identity:

\sin^2\theta = \frac{1 - \cos(2\theta)}{2}

U(t) = \frac{1}{4} k A^2 [1 - \cos(2 \times 2\pi n t)] = \frac{1}{4} k A^2 [1 - \cos(2\pi (2n) t)]

5 The oscillating term has frequency

2n

6 Match Option (2).

🛡️

Why You Can 100% Trust This Shortcut: Canonical harmonic oscillator trigonometric identity

\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Expressing Displacement in Simple Harmonic Motion

For a particle executing simple harmonic motion with frequency $n$ (angular frequency $\omega = 2\pi n$ ) and amplitude $A$ , displacement from the mean position is:

x(t) = A \sin(\omega t) = A \sin(2\pi n t)

Step 2

Formulating Potential Energy as a Function of Time

The potential energy $U$ of the harmonic oscillator with force constant $k$ is given by:

U(t) = \frac{1}{2} k x(t)^2 = \frac{1}{2} k A^2 \sin^2(\omega t)

Step 3

Applying the Power-Reduction Trigonometric Identity

Using the identity $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$ :

U(t) = \frac{1}{4} k A^2 \left[1 - \cos(2\omega t)\right] = \frac{1}{4} k A^2 - \frac{1}{4} k A^2 \cos(2\omega t)

Step 4

Determining the Frequency of the Oscillating Energy Term

The time-dependent harmonic term $\cos(2\omega t)$ has angular frequency $\omega' = 2\omega$ . Since $\omega = 2\pi n$ :

\omega' = 2(2\pi n) = 2\pi (2n) \implies \text{Frequency } f' = 2n

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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 1

✗ Incorrect

Verdict & Scientific Reason:

This is the frequency of displacement, velocity, and acceleration, but energy depends on the square of displacement ( $x^2$ ), doubling its frequency.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming energy oscillates at the same frequency as displacement.

2 2

✓ 100% Correct

Verdict & Scientific Reason:

Exact frequency of potential energy oscillations: $\omega' = 2\omega \implies f' = 2n$ .

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Why this is airtight: None; rigorous mathematical SHM analysis.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Unphysical harmonic multiplier.

⚠️

Student Fallacy (Why candidates pick this trap): Conflating third harmonics with quadratic functions.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Incorrect quadrupling of frequency.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing 4 quadrants of motion with frequency.

⚠️

NTA Trap Alert: Displacement Frequency vs. Energy Frequency Trap

Students frequently assume all parameters in SHM have frequency $n$ . Displacement, velocity, and acceleration have frequency $n$ . Kinetic energy and potential energy have frequency $2n$ . Total energy has frequency $0$ (constant).

🛡️ How to avoid the -1 mark: Remember:

x, v, a

oscillate at

n

. Kinetic energy and potential energy oscillate at

2n

. Total energy is constant (frequency = 0).

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Core NEET Syllabus Standard

NCERT Physics · Oscillations Standard Benchmark

Core Scientific Principle Tested:

"Both kinetic and potential energies of a particle in SHM vary periodically with twice the frequency of the displacement motion ( $f' = 2n$ )."

Syllabus Focus: Energy in Simple Harmonic Motion

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 27 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Double Frequency Rule: Energy Completes 2 Cycles per 1 SHM Cycle

Step-by-Step Rigorous Workflow & Mathematical Proof

Expressing Displacement in Simple Harmonic Motion

Formulating Potential Energy as a Function of Time

Applying the Power-Reduction Trigonometric Identity

Determining the Frequency of the Oscillating Energy Term

360° Option Elimination Engine Key Feature

NTA Trap Alert: Displacement Frequency vs. Energy Frequency Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Oscillations

Time Period of a Floating Cylindrical Cork in Liquids of Different Densities

Condition for a Circular Phase Space Trajectory in Simple Harmonic Motion

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

Ratio of Amplitudes for Identical Masses with Equal Maximum Speeds

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