A particle executes simple harmonic motion. Its amplitude is 8cm...

Given here, $A = 8cm$ and $T = 6s$

The displacement equation in SHM, when amplitude is halved is $Acos\left(\dfrac{2\pit}{T}\right) = \dfrac{A}{2}$

$⇒ \dfrac{2\pit}{T} = \dfrac{\pi}{3}$

Time taken by the particle is $t =$ $\dfrac{T}{6} = \dfrac{6}{6} = 1sec$.