If the period of SHM is 12 seconds and amplitude 10 cm

SOLUTION

The motion starts from extreme position, $y=A\cos \omega t=A\cos \left(\frac{2\pi }{T}\right)t$
$=10\cos [\left(\frac{2\pi }{12}\right)\times 14]=10\cos \frac{7\pi }{3}$
$=10\cos \frac{\pi }{3}=10\times \frac{1}{2}=5cm.$
Position from mean position is 5 cm hence the total displacement from the positive extreme position is 5 cm