1. Use the SHM relations for maximum velocity and acceleration

For simple harmonic motion, $v_{\max}=A\omega$ and $a_{\max}=A\omega^2$

Hence, $\frac{a_{\max}}{v_{\max}}=\frac{A\omega^2}{A\omega}=\omega$

Given: $\frac{a_{\max}}{v_{\max}}=10\ \text{s}^{-1}$ So, $\omega=10\ \text{rad/s}$

2. Use the displacement equation at $t=0$

Take the SHM displacement as $x=A\sin(\omega t+\phi)$

Given initial phase: $\phi=\frac{\pi}{4}$

At $t=0$, $x(0)=A\sin\phi=5$

So, $A\sin\frac{\pi}{4}=5$ $A\cdot \frac{1}{\sqrt{2}}=5$ $A=5\sqrt{2}\ \text{m}$

3. Find maximum acceleration

$a_{\max}=A\omega^2$ $a_{\max}=5\sqrt{2}\times (10)^2$ $a_{\max}=5\sqrt{2}\times 100$ $a_{\max}=500\sqrt{2}\ \text{m/s}^2$

4. Match with the options

The correct option is: $\boxed{500\sqrt{2}\ \text{m/s}^2}$ which is Option B.