(a) Use conservation of energy:
SPEmax = 1/2*k*xmax^2
KEmax = 1/2*m*vmax^2
Because energy is conserved:
SPEmax = KEmax
k*xmax^2 = m*vmax^2
vmax = xmax*sqrt(k/m)
(B) Call the 1.5 cm compression point, point B. Use conservation of energy to equate total energy at B to SPEmax.
SPEmax = KEb + SPEb
KEb = 1/2*m*vb^2
SPEb = 1/2*k*xb^2
1/2*k*xmax^2 = 1/2*m*vb^2 + 1/2*k*xb^2
Solve for vb:
vb = sqrt(k/m*(xmax^2 - xb^2))
= 21.15 cm/s
(C) Because of symmetry of the simple harmonic motion, the velocity at the compression you mention in part B is equal and opposite to this velocity in part C. Because we are only interested in the speed, forget about the sign. Thus vc = vb.
(D) Call point D, this point of interest.
vd = vmax/2
From part A, vmax = xmax*sqrt(k/m), therefore:
vd = 1/2*xmax*sqrt(k/m)
Conservation of energy from SPEmax point to point D:
1/2*k*xmax^2 = 1/2*k*xd^2 + 1/2*m*vd^2
solve for xd:
vd = ?sqrt(xmax^2 - m/k * vd^2)
Plug in known expression for vd and simplify:
xd = ?xmax*sqrt(3)/2
+-4*sqrt(3)/2 = 3.464 cm
Both the positive and negative values apply, because this occurs during both tension and compression.
Summary of resulting expressions:
A) vmax = xmax*sqrt(k/m)
B) vb = sqrt(k/m*(xmax^2 - xb^2))
C) vc = -sqrt(k/m*(xmax^2 - xc^2)), but you can forget about the negative sign because we were asked for "speed".
D) xd = ?xmax*sqrt(3)/2
xmax:=0.04 m; k:=19.2N/m; m:=0.59 kg; xb:=0.015 m; xc:= -0.015 m;