Difference between revisions of "Analysis of a simple harmonic oscillator"

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We developed a program that calculates the actual velocity of a mass on a linear spring.
 
<math>
 
<math>
 
\def\MYdef{\mathrel{\stackrel{\rm def}=}}
 
\def\MYdef{\mathrel{\stackrel{\rm def}=}}
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\def\){\right)}
 
\def\){\right)}
 
</math>
 
</math>
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The Hooke's law has the form:
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Here is the program showing the dependence of the speed of movement of the sinker on a linear spring. For starters to get this dependency.
 
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We write the Hooke's law:
 
 
::<math>
 
::<math>
 
F = - Cu,
 
F = - Cu,
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</math>
 
</math>
  
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Divide by <math>m</math> and multiply by <math>\dot u</math>:
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Dividing by <math>m</math> and multiplying by <math>\dot u</math> we arrive at:
 
::<math>
 
::<math>
 
\ddot u \dot u + \omega^2 \dot u u = 0, \quad \omega_0 \MYdef \sqrt{\frac{C}{m}},
 
\ddot u \dot u + \omega^2 \dot u u = 0, \quad \omega_0 \MYdef \sqrt{\frac{C}{m}},
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</math>
 
</math>
  
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We introduce the notation:
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By introducing notation
 
::<math>\frac{\dot u}{\omega} = y, \quad u = x</math>.
 
::<math>\frac{\dot u}{\omega} = y, \quad u = x</math>.
  
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We get the formula of the circle:
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We obtain:
 
<math> y^2 + x^2 = 0 </math>.
 
<math> y^2 + x^2 = 0 </math>.
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We get the formula of the circle:
 
  
 
{{#widget:Iframe |url=http://tm.spbstu.ru/htmlets/Tcvetkov/Spring/New_spring_v1.3_no_constructor/draw_spring.html |width=830 |height=550 |border=0 }}
 
{{#widget:Iframe |url=http://tm.spbstu.ru/htmlets/Tcvetkov/Spring/New_spring_v1.3_no_constructor/draw_spring.html |width=830 |height=550 |border=0 }}
  
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Разработчики: [[D.V. Tsvetkov]], [[A.M. Krivtsov]].
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Developers: [[D.V. Tsvetkov]], [[A.M. Krivtsov]].
  
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[[Category: Virtual laboratory]]
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[[Category: Programming]]
 
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Latest revision as of 18:31, 18 January 2017

We developed a program that calculates the actual velocity of a mass on a linear spring. [math] \def\MYdef{\mathrel{\stackrel{\rm def}=}} \def\({\left(} \def\){\right)} [/math] The Hooke's law has the form:

[math] F = - Cu, [/math]
[math] m \ddot u + Cu = 0. [/math]

Dividing by [math]m[/math] and multiplying by [math]\dot u[/math] we arrive at:

[math] \ddot u \dot u + \omega^2 \dot u u = 0, \quad \omega_0 \MYdef \sqrt{\frac{C}{m}}, [/math]
[math] \dot u^2 + \omega^2 u^2 = 0, \quad \(x^2(t)\)'_t = 2x(t)\dot u(t), [/math]
[math] \(\frac{\dot u}{\omega}\)^2 + u^2 = 0. [/math]

By introducing notation

[math]\frac{\dot u}{\omega} = y, \quad u = x[/math].

We obtain: [math] y^2 + x^2 = 0 [/math].

Developers: D.V. Tsvetkov, A.M. Krivtsov.


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