Answer
(b) $\gt$ (c) $\gt$ (a)
Or
$A_b \gt A_c \gt A_a$
Work Step by Step
We have
$\frac{1}{2}kA² = \frac{1}{2}kx²+ \frac{1}{2}mv² $
Let $x_0$ be the streched length then for
(a) $\frac{1}{2}kA² = \frac{1}{2}k{x_0}²$
$A_a= x_0 $
(b) $\frac{1}{2}kA² = \frac{1}{2}kx²+ \frac{1}{2}mv² $
$\frac{1}{2}kA² = \frac{1}{2}kx_0²+ \frac{1}{2}mv_0² $
$A_b = \sqrt{x_0² + (m/k )v_0²}$
(c) $\frac{1}{2}kA² = \frac{1}{2}kx²+ \frac{1}{2}mv² $
$\frac{1}{2}kA² = \frac{1}{2}kx_0²+ \frac{1}{2}m(\frac{1}{2}v_0² )$
$A_c = \sqrt{x_0² + (m/2k)v_0²}$
From the amplitudes we saw
$A_b \gt A_c \gt A_a$
