A particle executing simple harmonic motion has a time period of 4...

Given, time period $T = 4 \text{ s} and amplitude$A = 6 \text{ m}$.<br><br>The angular frequency$\omega$is given by:<br>$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \text{ rad/s}$<br><br>The general equation for the displacement of a particle in SHM is:<br>$x = A \sin(\omega t + \phi)$<br><br>Since the particle starts from the positive extreme position at$t = 0$, its initial phase$\phi = \dfrac{\pi}{2}$.<br><br>Substituting the values into the general equation:<br>$x = 6 \sin\left(\dfrac{\pi t}{2} + \dfrac{\pi}{2}\right)$<br><br>Using the trigonometric identity$\sin\left(\theta + \dfrac{\pi}{2}\right) = \cos(\theta)$, we get:<br>$x = 6 \cos\left(\dfrac{\pi t}{2}\right)$<br><br>Answer:$x = 6 \cos\left(\dfrac{\pi t}{2}\right)