Total energy = K.E. + P.E. at x=0.04 m, T.E. =0.5+0.4=0.9 J T.E ⇒A A=1 mω2 A2=0.9=21×0.2(2π×π25)2×A2=0.9=0.06 m=6 cm
JEE Main 2024 — Physics Waves & Oscillations
An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (π25)Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is _____cm.
Held on 8 Apr 2024 · Verified 6 Jul 2026.
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