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2014-08-19T01:05:32-03:00
\boxed{\boxed{y(x,t)=(6,00mm)*sin(kx+(600rad/s))*t+\phi}}

iremos achar um intervalo de tempo \boxed{t_2 - t_1}

t1 = será o tempo quando y= +2mm
t2 = será o tempo quando y= -2mm
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observando a equação da origem e sem deslocamento de onda
x=0\\\phi =0

y(0,t)=(6,00mm)*sin(k*0x+(600rad/s)*t+0 \\\\y(t)=6,00mm*sin(600rad/s)*t
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quando y=+2...temos o t1


2,0mm=6,00mm*sin(600rad/s)*t_1\\\\ \frac{2,0mm}{6,0mm} =sin(600rad/s)*t_1\\\\arcsen(\frac{2,0mm}{6,0mm})=(600rad/s)*t_1\\\\\\\boxed{\boxed{ \frac{\arcsen(\frac{2,0mm}{6,0mm})}{600rad/s} =t_1}}

aplicando o mesmo para t2
que será quando y=-2 teremos

{\boxed{ \frac{\arcsen(\frac{-2,0mm}{6,0mm})}{600rad/s} =t_2}}

agora fazendo t1 - t2 = T 
T = intervalo de tempo

t1-t2=T \\\\ \frac{arcsen(\frac{2,0mm}{6,0mm})}{600rad/s} -\frac{ arcsen(\frac{-2,0mm}{6,0mm})}{600rad/s} =T\\\\ \frac{arcsen(\frac{2,0mm}{6,0mm})}{600rad/s} +\frac{ arcsen(\frac{2,0mm}{6,0mm})}{600rad/s} =T\\\\\\ \frac{2*(arcsen( \frac{2,00)}{6,00} )}{600rad/s} =T\\\\\boxed{1,13*10^{-3} s=T}

resposta
1,1 m/s = intervalo de tempo



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