The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s$$^{-1}$$. At, $$t = 0$$ the displacement is 5 m. What is the maximum acceleration? The initial phase is $$\frac{\pi}{4}$$.

For a particle in simple harmonic motion we always describe its displacement by the standard equation

$$x(t)=A\sin\!\left(\omega t+\phi\right)$$

where $$A$$ is the amplitude, $$\omega$$ is the angular frequency and $$\phi$$ is the initial phase.

From this displacement we obtain the velocity by differentiating once with respect to time. Using the derivative of sine, $$\dfrac{d}{dt}\bigl[\sin(\theta)\bigr]=\omega\cos(\theta)$$, we write

$$v(t)=\dfrac{dx}{dt}=A\omega\cos\!\left(\omega t+\phi\right).$$

The maximum value of the cosine function is $$+1$$, so the maximum (positive) speed is

$$v_{\text{max}}=A\omega.$$

Differentiating the velocity once more gives the acceleration. Using $$\dfrac{d}{dt}\bigl[\cos(\theta)\bigr]=-\omega\sin(\theta)$$, we get

$$a(t)=\dfrac{dv}{dt}=-A\omega^{2}\sin\!\left(\omega t+\phi\right).$$

The sine function again reaches an extreme value of $$\pm1$$, hence the magnitude of the maximum acceleration is

$$a_{\text{max}}=A\omega^{2}.$$

We are told that the ratio of maximum acceleration to maximum velocity equals $$10~\text{s}^{-1}$$. Writing this ratio explicitly and substituting the two expressions we have just obtained, we find

$$\frac{a_{\text{max}}}{v_{\text{max}}}=\frac{A\omega^{2}}{A\omega}=\omega.$$

Hence

$$\omega=10~\text{s}^{-1}.$$

Next, the displacement at time $$t=0$$ is given as $$x(0)=5~\text{m}$$ and the initial phase is specified as $$\phi=\dfrac{\pi}{4}$$. Substituting $$t=0$$ and $$\phi$$ into the displacement equation we write

$$x(0)=A\sin\!\left(\omega\cdot0+\frac{\pi}{4}\right)=A\sin\!\left(\frac{\pi}{4}\right).$$

We know that $$\sin\!\left(\frac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$$, so

$$5=\frac{A}{\sqrt{2}}\quad\Longrightarrow\quad A=5\sqrt{2}\;\text{m}.$$

Now we have both $$A$$ and $$\omega$$, so we can calculate the maximum acceleration using $$a_{\text{max}}=A\omega^{2}$$. Substituting the obtained values,

$$a_{\text{max}}=\bigl(5\sqrt{2}\bigr)\times(10)^{2}=5\sqrt{2}\times100=500\sqrt{2}~\text{m s}^{-2}.$$

Hence, the correct answer is Option D.