Precalculus (6th Edition) Blitzer

$8 \sin(\dfrac{\pi}{2}) t$
Amplitude, $A=|8|=8$ Now, $f=\dfrac{\omega}{2 \pi} \implies \omega =2 \pi f$ and period, $P=\dfrac{2 \pi}{\omega}$ So, $\omega =2 \pi (\dfrac{1}{4})=\dfrac{\pi}{2}$ Therefore, $d=A \sin \omega t=8 \sin(\dfrac{\pi}{2}) t$