Physics

Physics

Problem

For a particle of mass mm executing SHM with angular frequency ω\omega and amplitude AA, which expression gives its kinetic energy KK when its instantaneous displacement from equilibrium is xx? a) 12mω2x2\dfrac{1}{2} m\omega^{2}x^{2} b) mω2(A2−x2)m\omega^{2}\bigl(A^{2}-x^{2}\bigr) c) 12mω2(A2−x2)\dfrac{1}{2} m\omega^{2}\bigl(A^{2}-x^{2}\bigr) d) 12mω2(A2+x2)\dfrac{1}{2} m\omega^{2}\bigl(A^{2}+x^{2}\bigr)
Intuitive explanation

The key is recognizing that energy in SHM converts between kinetic and potential forms while staying constant overall. Since the total mechanical energy is E=12mω2A2E = \frac{1}{2}m\omega^2A^2 (maximum potential energy at amplitude), you can find kinetic energy at position xx by subtracting the potential energy U=12mω2x2U = \frac{1}{2}m\omega^2x^2 from the total. This gives you kinetic energy as a function of position without needing to find velocity explicitly.

Intuitive explanation

The key is recognizing that energy in SHM converts between kinetic and potential forms while staying constant overall. Since the total mechanical energy is E=12mω2A2E = \frac{1}{2}m\omega^2A^2 (maximum potential energy at amplitude), you can find kinetic energy at position xx by subtracting the potential energy U=12mω2x2U = \frac{1}{2}m\omega^2x^2 from the total. This gives you kinetic energy as a function of position without needing to find velocity explicitly.

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