16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4d4,{"@context":"https://schema.org","@type":"Quiz","name":"Viteee 2024 Simple Harmonic Motion PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/viteee/questions/2024/physics/simple-harmonic-motion/m9bd69ly","educationalLevel":"Entrance exam preparation","about":"Viteee / Physics / Simple Harmonic Motion","hasPart":{"@type":"Question","name":"Viteee 2024 Simple Harmonic Motion PYQ with Solution","text":"The potential energy of a particle of mass 0.1 kg moving along the X-axis, is given by U=5 x(x-4) \\mathrm{J}, where x is in metres. Choose the wrong option.","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"D"},"comment":{"@type":"Comment","text":"The potential energy U of a particle with mass 0.1 kg moving along the X -axis is described by: U = 5x(x-4) \\, \\text{J} where x is the position in meters. The equation expands to: U = 5x^2 - 20x To find the force F , we take the derivative of U with respect to x : \\frac{dU}{dx} = 10x - 20 The force is given by the negative derivative of the potential energy: F = -\\frac{dU}{dx} = -10x + 20 Assuming a shifted position x = (X + 2) , the force becomes: F = -10X The force is zero when X = 0 or equiv..."}}}18:T4ce,

The potential energy $ U $ of a particle with mass 0.1 kg moving along the $ X $-axis is described by:

$ U = 5x(x-4) \, \text{J} $

where $ x $ is the position in meters.

The equation expands to:

$ U = 5x^2 - 20x $

To find the force $ F $, we take t

Simple Harmonic Motion

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