For 1st qn,
A = 3
w = 2pi / 2
= pi
x = A sin (wt) only when t = 0 and x = 0
Since the period is 2 seconds, it takes 1/2 second for the mass to
travel to the rest position due to SHM motion.
After 0.75 second, t= 0.75 - 0.5 = 0.25
t = 0.25
x = 2.12
Therefore, displacement of the mass from rest position after 0.75
sec is 2.12 cm.
After 1 sec, the mass reaches the lowest point if the mass is
displaced upwards earlier on. Thus, there is no overlapse between
distance and displacement.
Therefore,
Total distance travelled = 3.0 + 2.12
= 5.12
= 5.1 cm
For 2nd qn,
U = 1/2 mv(max)^2
1.0 = 1/2(4)(v[max])^2
0.5 = v(max)^2
v(max) = root (2)/2
v(max) = Aw
= 0.2 (2pi/T)
square root (2)/2 = 0.4pi / T
T = 4 pi / (5[square root {2} ] )