Distance of particle from mean position, x1=1cm Acceleration, a=3cm∕s2 Velocity, v2=6cm∕s, when x2=2cm. As we know that, v2=ω√A2−x22 ∴6=ω√A2−22 On squaring both sides, we get 62=ω2(A2−4) . . . (i) and since, a=−ω2x where, a is acceleration. ∴3=ω2 ⇒ω2=3 put in Eq. (i), we get 36=3(A2−4) ⇒12=A2−4 ⇒A2=12+4=16 ⇒A=4cm
