A body executing SHM has a maximum velocity 1 m/sec, maximum acceleration of $4 m/se{c}^2$. It’s amplitude in meters is

$$\begin{gathered} Given velocity maximum is {\text{v}}_{\max }=1\text{m/s=ωA, here w=angular velocity,A=amplitude} \\ acceleration maximum {\text{a}}_{\text{max}}{\text{=4m/s}}^{\text{2}}{\text{=ω}}^2\text{A} \\ divide above two equations \omega {\text{=a}}_{\text{max}}{\text{/v}}_{\text{max}}=4 \\ we know that, {\text{v}}_{\text{max}}=\text{A×ω} \\ \Rightarrow \text{A=}\frac{{\text{v}}_{\text{max}}}{\text{ω}}\text{, here A=amplitude} \\ \Rightarrow A=\frac{1}{4}=0.25 m \end{gathered}$$