JEE MAIN - Physics (2002 - No. 40)
In a simple harmonic oscillator, at the mean position
kinetic energy is minimum, potential energy is maximum
both kinetic and potential energies are maximum
kinetic energy is maximum, potential energy is minimum
both kinetic and potential energies are minimum.
Forklaring
$$K.E = {1 \over 2}k\left( {{A^2} - {x^2}} \right);\,\,\,U = {1 \over 2}k{x^2}$$
At the mean position $$x=0$$
$$\therefore$$ $$K.E. = {1 \over 2}k{A^2} = $$ Maximum and $$U=0$$
At the mean position $$x=0$$
$$\therefore$$ $$K.E. = {1 \over 2}k{A^2} = $$ Maximum and $$U=0$$