Two strings are adjusted to vibrate at exactly 420 Hz . Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

423 Hz

Explanation:

The frequency of a string is given by

k is a constant representing the mode of oscillation, l is the length of the string, T is the tension and m is linear density of the string.

When other factors are constant, it is seen that as the tension increases, the frequency increases. So it is expected that the tightened string will have a higher frequency.

The beat frequency is the difference in the frequencies of both string. From the question, this beat frequency is 3 Hz. Hence, the tightened string, having a highrt frequency, has a frequency of 420 + 3 Hz = 423 Hz

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

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.