Exploring Displacement

$x = A\cos\theta$ (correct)

$x = A\sin\theta$

$x = A\sec\theta$

$x = A\tan\theta$

$a = \omega^2A^2$

$a = \omega^2A$ (correct)

$a = \omegaA$

$a = \omegaA^2$

$a = -\omega^2A\tan\theta$

$a = -\omega^2A\cos\theta$ (correct)

$a = -\omega^2A\sin\theta$

$a = -\omega^2A\sec\theta$

<p>The acceleration of point P is directly proportional to its displacement x.</p> Signup and view all the answers

<p>$\omega = \frac{2\pi}{T}$</p> Signup and view all the answers