Question 178: Oscillations

Question

In a simple harmonic oscillator, at the mean position

Accepted Answer

$$K.E = {1 \over 2}k\left( {{A^2} - {x^2}} ight);\,\,\,U = {1 \over 2}k{x^2}$$
At the mean position
$$x=0$$
$$	herefore$$  $$K.E. = {1 \over 2}k{A^2} = $$ Maximum and $$U=0$$

Answer

kinetic energy is minimum, potential energy is maximum

Answer

both kinetic and potential energies are maximum

Answer

kinetic energy is maximum, potential energy is minimum

Answer

both kinetic and potential energies are minimum.

Oscillations

In a simple harmonic oscillator, at the mean position

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