T=2πkm
0.2=2πk0.5
k=50π2
≈500
x=Asin(ωt+ϕ)
=5cmsin(4ωT+0)
=5cmsin(2π)
=5cm
PE=21kx2
=21(500)(1005)2
=0.6255
JEE Main 2021 — Physics Waves & Oscillations
An object of mass 0.5kg is executing simple harmonic motion. It amplitude is 5cm and time period (T) is 0.2s. What will be the potential energy of the object at an instant t=4Ts starting from mean position. Assume that the initial phase of the oscillation is zero.
Held on 27 Jul 2021 · Verified 6 Jul 2026.
0.62J
6.2×10−3J
1.2×103J
6.2×103J
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