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Harmoniske svingninger
Posted: 05/10-2021 12:43
by Eddie
Noen tips for hvordan man løser denne oppgaven?

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Re: Harmoniske svingninger
Posted: 05/10-2021 14:10
by Mattebruker
Gitt
N[tex]_{h}[/tex]( t ) = 1200 - 300 ( cos([tex]\frac{\pi }{3}[/tex]t ) +[tex]\sqrt{3}\cdot[/tex]sin([tex]\frac{\pi }{3}[/tex]t ) )
= 1200 - 300[tex]\cdot[/tex]2( [tex]\frac{1}{2}[/tex][tex]\cdot[/tex]cos([tex]\frac{\pi }{3}[/tex] t) + [tex]\frac{\sqrt{3}}{2}[/tex][tex]\cdot[/tex] sin([tex]\frac{\pi }{3}[/tex] t ) )
= 1200 - 600 ( cos[tex]\frac{\pi }{3}[/tex][tex]\cdot[/tex]cos([tex]\frac{\pi }{3}[/tex] t ) + sin([tex]\frac{\pi }{3}[/tex]) [tex]\cdot[/tex]sin([tex]\frac{\pi }{3}[/tex]t ) )
[ cosu[tex]\cdot[/tex]cosv + sinu[tex]\cdot[/tex]sinv = cos(u - v) ]
= 1200 - 600 [tex]\cdot[/tex] cos([tex]\frac{\pi }{3}[/tex]t - [tex]\frac{\pi }{3}[/tex] )
= 1200 - 600 [tex]\cdot[/tex]cos[tex]\frac{\pi }{3}[/tex]( t - 1 )
[ bølgetalet k = [tex]\frac{\pi }{3}[/tex] = [tex]\frac{2\pi }{T}[/tex] [tex]\Leftrightarrow[/tex] perioden T = 6 ]
= 1200 - 600 [tex]\cdot[/tex] cos([tex]\frac{2\pi }{6}[/tex]( t - 1 ))
Re: Harmoniske svingninger
Posted: 05/10-2021 15:32
by Eddie
Wow, tusen takk!!!