Question

An object moves in simple harmonic motion with a period of 8 seconds and amplitude of 6 cm. At time t = 0 seconds, its displacement d from rest is 0 cm, and initially, it moves in a negative direction.

a) True
b) False

Answer 1

The total distance moved by a particle in one time period of simple harmonic motion is 4 times the amplitude (4A), which accounts for a complete cycle of back-and-forth motion.

Step-by-step explanation:

The student's question is about simple harmonic motion (SHM) which involves an object moving back and forth in a periodic manner.

In the context of the question, defining distance moved by a particle in one time period in SHM would mean considering the path traveled during one complete cycle of motion. Given that the amplitude A represents the maximum displacement from the equilibrium point, the distance covered in one full cycle is not simply A cm, as that would only represent the displacement in one direction. Instead, since the object travels from the equilibrium point to the amplitude A in one direction, and then back past equilibrium to the amplitude A on the other side, the total distance covered in one time period would be 4A cm.

The object would move from its maximum positive displacement to the maximum negative displacement, covering 2A, and then back to its maximum positive displacement, covering another 2A. Therefore, the correct answer to the question would be 4A cm, which isn't one of the given options in the question.