In simple harmonic motion, the speed is greatest at that point in the cycle whena) the magnitude of the acceleration is a maximum. b) the displacement is a maximum. c) the magnitude of the acceleration is a minimum. d) the potential energy is a maximum. e) the kinetic energy is a minimum.

C) the magnitude of the acceleration is a minimum.

Explanation:

As we know that ,the general equation of the simple harmonic motion given as

The displacement x given as

x=X sinω t

Then the velocity v will become

v= X ω cosωt

The acceleration a

a= - X ω² sinω t

The speed of the particle will be maximum when cosωt will become 1 unit.

It means that sinωt will become zero.So acceleration and displacement will be minimum.

Therefore when speed is maximum then acceleration will be minimum.

At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.

Therefore option C is correct.

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.

i think it would be 5