Question:

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement

Updated On: Dec 9, 2024
  • 2A2\sqrt{A}
  • A2\frac{A}{2}
  • A2\frac{A}{\sqrt{2}}
  • A2A\sqrt{2}
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The Correct Option is C

Solution and Explanation

In simple harmonic motion (SHM), the total energy (E) is constant and given by E = PE + KE where, PE is potential energy and KE is kinetic energy.

At any displacement (x) from the equilibrium position:

PE = 12kx2\frac{1}{2}kx^2

KE = 12k(A2x2)\frac{1}{2}k(A^2 - x^2)

When PE = KE: 12kx2=12k(A2x2)\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)

x2=A2x2x^2 = A^2 - x^2

2x2=A22x^2 = A^2

x=A2x = \frac{A}{\sqrt{2}}

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