Acceleration in terms of x (1 Viewer)

Catz · Oct 25, 2015

Express this x=cos^2(3t) in acceleration in terms of x

InteGrand · Oct 25, 2015

Catz said:
Express this x=cos^2(3t) in acceleration in terms of x

$Using the trigonometric identity $\cos ^2 \theta = \frac{1}{2} + \frac{1}{2}\cos 2\theta$, we have$
$\begin{align*}x &= \cos^2 3t \\ \Rightarrow x&= \frac{1}{2} + \frac{1}{2} \cos 6t \\ \Rightarrow x-\frac{1}{2} &= \frac{1}{2}\cos 6t. \end{align*}$

$This is of the form $x-x_0 = A\cos nt + B\sin nt$, which is simple harmonic motion (S.H.M.) and thus satisfies the S.H.M. differential equation $\ddot{x} = -n^2 (x-x_0)$ ($x_0$ is the centre of motion). In our case, we have $x_0 =\frac{1}{2}$ and $n =6\Rightarrow n^2 =36$, so we have $\ddot{x} = -36\left(x-\frac{1}{2}\right).$

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Acceleration in terms of x (1 Viewer)

Catz

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InteGrand

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