Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 639: 62

$d=\sin 196.2\times10^6 \pi t $

Amplitude, $ A=|1|=1$ Now, $ f=\dfrac{\omega}{2 \pi} \implies \omega =2 \pi f $ and period, $ P=\dfrac{2 \pi}{\omega} $ So, $\omega =2 \pi (98.1\times10^6)=196.2\times10^6 \pi $ Therefore, $ d=A \sin \omega t= \sin 196.2\times10^6 \pi t $