A) a 32-pound weight stretches a spring 2 feet. determine the amplitude and period of motion if the weight is released 1 foot above the equilibrium position with an initial upward velocity of 2 ft/sec. (b) how many complete vibrations will the weight have completed at the end of 4π seconds?

Explanation:

Value of spring constant K = weight / extension

=

=512 ft s⁻²

Frequency of oscillation of spring mass system

=

Putting the values we get

frequency

=" alt="\sqrt{\frac{512}{32} }" />" />

= 4  .

Time period = 1 / frequency = .25 s

No of vibrations in 4π second

=

= 50 approx

When the weight is released 1 foot above equilibrium

PE stored = 1/2 K x² = .5 x 512 x 1 = 256 J

KE = 1/2 mv² = .5 x 32 x 4 = 64 J

Total energy =  320 J

Let x be the amplitude upto which spring stretches due to this energy

1/2 k x² = 320 ,

1/2 x 512  x² = 320

x = 1.12 ft approx

c) up

hope this

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.