A body executing SHM has a maximum acceleration equal to $48 m/se{c}^2$ and maximum velocity equal to 12 m/sec. The amplitude of SHM is

given maximum acceleration ${\text{a}}_{\text{max}}{\text{=48m/s}}^{\text{2}}$

Maximum velocity  ${\text{V}}_{\text{max}}\text{=12m/s}$
amplitude A=? here angular velocity =$\omega$
$\frac{{\text{a}}_{\text{max}}}{{\text{V}}_{\text{max}}}=\frac{{\text{ω}}^{\text{2}}\text{A}}{\text{ωA}}=\frac{48}{12}=4$
=> $\omega$=4 
${\text{V}}_{\text{max}}=\text{ωA}$

substitute maximum velocity, and angular velocity
12=4(A)
A=3m.