The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=A \sin \omega t, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T \/ 2 \beta. The value of \beta is \_\_\_\_ . [28 Jan 2026 Shift 1]

JEE Main 28 Jan 2026 Shift 1 Paper Solved Paper

Section: Physics

Question : 47 of 75

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The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t)=Asin‌ωt, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T∕2β. The value of β is ____ .

[28 Jan 2026 Shift 1]

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