User:Krauss/Sandbox

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[edit] TESTE HIDEN

Equivalent Spring Constant (Series)

Deriving keq in the series case is a little trickier than in the parallel case. Defining the equilibrium position of the block to be x2, we'll be looking for equation for the force on the block that looks like:

F_b = -k_{eq} x_2 .\,

To begin, we'll also define the equilibrium position of the point between the two springs to be x1.

The force on the block is

F_b = - k_2 \left( x_2 - x_1 \right). \quad \quad \quad (1) \,

Meanwhile, the force on the point between the two springs is

F_s = - k_1 x_1 + k_2 (x_2 - x_1). \,

Now, when the block is pushed so the springs are compressed and the system is allowed to come to equilibrium, the force between the strings must sum to zero, so with Fs = 0 we can solve for x_1 \,:

- k_1 x_1 + k_2 (x_2 - x_1) = 0 \,
- k_1 x_1 - k_2 x_1 = -k_2 x_2 \,
\left(k_1 + k_2 \right) x_1 = k_2 x_2 \,

so

x_1 = \frac{k_2}{k_1 + k_2} x_2 . \,

Now we just plug this back into (1):

F_b \, = -k_2 x_2 + k_2 x_1 \,
= -k_2 x_2 + k_2 \left( \frac{k_2}{k_1 + k_2} x_2 \right) \,
= -k_2 x_2 \left( \frac{k_1 + k_2}{k_1 + k_2} \right) + \frac{k_2^2}{k_1 + k_2} x_2 \,
= x_2 \frac{-k_1 k_2 - k_2^2 + k_2^2}{k_1 + k_2} \,

Finally, the force on the block has been found:

F_b = - \left( \frac{k_1 k_2 }{k_1 + k_2} \right) x_2 .\,

So we can define everything in the parenthesis to be

k_{eq} = \frac{k_1 k_2 }{k_1 + k_2} .\,

Which can also be written:

\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}. \,


[edit] Test2

User:Krauss/WTF

ver tb Template_system_formalism/FAQ... Web_template_system_(formalism)

On CPAN's Perl code, all pl file is documented at the "__END__" mark. But there are another template language, it look like this:

... perl code ...
__END__

=head1 NAME

ProgramName - description...

=head1 SYNOPSIS
...

Nononono no no non onononono n nononononno User:Krauss/ART nononon onononononon no nonnonononononono nonononono.


Lembrete: Web_template_hook_styles