7.4.35 | до правки #12882 | Решения Савченко

Statement

7.4.35^∗. What is the period of small vibrations of four charged bodies connected by
identical filaments of length l and moving as shown in the figure? Mass and
charge of the body m and q.

Solution

The First Way

Force Way

The diagonal
$d=l\times\sqrt(2)$
$$$d1=d+x$$$
There are 2 ways to find out difference of force
first
$$$dF=F2-F1=k\timesq\timesq\times(1/(d\timesd)-1(d1\timesd1))$$$
The second way
$$$F=k\timesq\timesq/(d1\timesd1)$$$
$dF/dx=-2k×q×q/(d-x)\wedge3$

k' is stifness of spring

$$$k'\timesdx=dF$$$
$k'=2×k×q×q×dx/d\wedge3$
But the spring located between 2 object.We need period one of them

The period formula
$T=2×\Pi×\sqrt(m/K)$
But the spring is between 2 object
We need use half of spring
so
$$$K=k'\times2$$$
The period is
$T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q))$

#### Answer

$T=2×\Pi×\sqrt(4×\Pi×\varepsilon×m×l×l/\sqrt(2)×q×q)$