The Question is from neet-physics oscillations-waves energy-of-a-particle-executing-shm, and it was created by the author P.NARASIMHA

Question

A particle of mass $m$ executes SHM with amplitude $A$ and angular frequency $\omega$. What is the average kinetic energy of the particle over one complete oscillation?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION ✓ **Answer**: B ## CONCEPT FRAMEWORK **Primary Concept**: Average Kinetic Energy in Simple Harmonic Motion **Related Topics**: Simple Harmonic Motion, Kinetic Energy, Potential Energy, Energy Conservation **Prerequisites**: Understanding of SHM, kinetic energy formula, integration ## SOLUTION PATHWAY **Given Information**: Mass of particle = $m$ Amplitude of SHM = $A$ Angular frequency = $\omega$ **Method & Steps**: 1. Express the velocity of the particle in SHM. The displacement of a particle in SHM is given by $x(t) = A\sin(\omega t)$. The velocity is the derivative of displacement with respect to time: $v(t) = \frac{dx}{dt} = A\omega\cos(\omega t)$ 2. Calculate the kinetic energy. Kinetic energy, $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(A\omega\cos(\omega t))^2 = \frac{1}{2}m A^2\omega^2\cos^2(\omega t)$ 3. Find the average kinetic energy over one complete oscillation. The average value of $\cos^2(\omega t)$ over one complete oscillation ($0 \le \omega t \le 2\pi$) is $\frac{1}{2}$. Therefore, the average kinetic energy is: $ = \frac{1}{2}m A^2\omega^2 imes \frac{1}{2} = \frac{1}{4}m A^2\omega^2$ Alternatively, using integration: $ = \frac{1}{T}\int_0^T \frac{1}{2}m A^2\omega^2\cos^2(\omega t) dt = \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{2}m A^2\omega^2\cos^2( heta) d heta = \frac{1}{4}m A^2\omega^2$ Where $T = \frac{2\pi}{\omega}$ is the period of oscillation and $ heta = \omega t$ 4. Result: Average kinetic energy $= \frac{1}{4}m\omega^2 A^2$ Verification: The average kinetic energy should be half the total energy in SHM. The total energy is $\frac{1}{2}m\omega^2 A^2$, and half of this is indeed $\frac{1}{4}m\omega^2 A^2$. ## OPTION ANALYSIS A :x: This is the total energy of the particle in SHM, not the average kinetic energy. B ✓: This is the correct average kinetic energy. C :x: This is twice the total energy of the particle in SHM. D :x: The average kinetic energy is non-zero; it's half the total energy. ## LEARNING AIDS **Common Mistakes**: 1.Confusing average kinetic energy with total energy: Remember that the average kinetic energy is half the total energy in SHM. 2.Incorrectly averaging $\cos(\omega t)$ instead of $\cos^2(\omega t)$: The average of $\cos(\omega t)$ over a complete cycle is zero, but the average of $\cos^2(\omega t)$ is $\frac{1}{2}$. **Quick Tips**: 1.Remember the quick formula: Average KE = $\frac{1}{4}m\omega^2 A^2$ for SHM. 2.Relate the average KE to the total energy in SHM: Average KE = Total Energy / 2

P.NARASIMHA · Answer

$\frac{1}{2} m \omega^2 A^2$

P.NARASIMHA · Answer

The correct Option is "$\frac{1}{4} m \omega^2 A^2$"

P.NARASIMHA · Answer

$m \omega^2 A^2$

P.NARASIMHA · Answer

Zero

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