E=21Ka2
43E=21K(a2−y2)v
43×21Ka2=21K(a2−y2)
y2=a2−43a2
y=2a
A particle starts executing simple harmonic motion (SHM) of amplitude a and total energy E. At any instant, its kinetic energy is 43E, then its displacement y is given by:
Held on 27 Jul 2021 · Verified 6 Jul 2026.
y=a
y=2a
y=2a3
y=2a
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