For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy

Question

For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy (A) twice during each cycle (B) four times during each cycle (C) when x=0 (D) when x=a

Tardigrade Admin · Accepted Answer

Total energy = P E + K E . When a particle executes SHM, there is one position between maximum amplitude and mean position where PE is equal to KE , and particle have to cross that point four times to complete one oscillation.

Q. For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy

twice during each cycle

four times during each cycle

when $x = 0$

when $x = a$

Total energy = $P E + K E$ .
When a particle executes SHM, there is one position between maximum amplitude and mean position where $PE$ is equal to $K E$ , and particle have to cross that point four times to complete one oscillation.

Q. For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy

twice during each cycle

four times during each cycle

when x=0

when x=a

Total energy = PE+KE . When a particle executes SHM, there is one position between maximum amplitude and mean position where PE is equal to KE , and particle have to cross that point four times to complete one oscillation.

when $x = 0$

when $x = a$

Total energy = $P E + K E$ .
When a particle executes SHM, there is one position between maximum amplitude and mean position where $PE$ is equal to $K E$ , and particle have to cross that point four times to complete one oscillation.