In principle, the motion of a simple pendulum and an object on an ideal spring can both be used to provide the basic time interval or period used in a clock. Which of the two kinds of clocks becomes more inaccurate when carried to the top of a high mountain?

Simple-pendulum clock

Explanation:

A simple pendulum clock is defined as a clock that is comprised of a fixed mass, that is hanged at one end, which is commonly known as the pivot with the help of a string that is comprised of a definite length. This pendulum is then allowed to swing back and forth by simply displacing it at a certain angle and thereby releasing it. This results in the periodic motion.

The time period of this type of clock directly depends on its rate of acceleration under the influence of gravitational force. Due to this, it is considered to be an inaccurate clock when used at the top of high mountains.

magnified images possible images can be real = projected onto a screen

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.