$\displaystyle \frac{d^2x}{dt^2} + \frac{5k}{m} \cdot x = 0 $ um... is this correct? $\displaystyle x(t) = B~cos(\sqrt5\frac{k}{m})+C~sin(\sqrt5\frac{k}{m}) $ or just... wrong...
Follow Math Help Forum on Facebook and Google+
Originally Posted by billym $\displaystyle \frac{d^2x}{dt^2} + \frac{5k}{m} \cdot x = 0 $ um... is this correct? $\displaystyle x(t) = B~cos(\sqrt5\frac{k}{m})+C~sin(\sqrt5\frac{k}{m}) $ or just... wrong... If you mean $\displaystyle x(t) = B~cos \left ( t \sqrt{\frac{5k}{m}} \right ) + C~cos \left ( t \sqrt{\frac{5k}{m}} \right )$ then yes, it is correct. -Dan
View Tag Cloud