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

Maximum velocity in SHM, ${\mathrm{\nu }}_{\max }=\mathrm{a\omega }$

Maximum acceleration in SHM, ${\mathrm{A}}_{\max }={\mathrm{a\omega }}^2$

Where a and $\mathrm{\omega }$ are amplitude and angular frequency.

Given that, $\frac{{\mathrm{A}}_{\max }}{{\mathrm{v}}_{\max }}=10 \mathrm{i}.\mathrm{e}., \mathrm{\omega }=10 {\mathrm{s}}^{-1}$

displacement is given by $\mathrm{x}=\mathrm{a} \sin \left(\mathrm{\omega t}+\mathrm{\pi }/4\right) \mathrm{at} \mathrm{t}=0, \mathrm{x}=5$

$$\begin{gathered} 5=\mathrm{a} \sin \mathrm{\pi }/4 \\ 5=\mathrm{a} \sin {45}^0\Rightarrow \mathrm{a}=5\sqrt{2} \\ \mathrm{Maximum} \mathrm{acceleration} {\mathrm{A}}_{\max }={\mathrm{a\omega }}^2=500\sqrt{2} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$