Answer
The new amplitude is $~~0.18~m$
Work Step by Step
We can use the original period to find the mass of the block:
$T = 2\pi \sqrt{\frac{m}{k}}$
$T^2 = \frac{4\pi^2 m}{k}$
$m = \frac{T^2~k}{4\pi^2}$
$m = \frac{(0.40~s)^2 (600~N/m)}{4\pi^2}$
$m = 2.43~kg$
Suppose the original maximum speed of the block was $v_1$
We can find the new maximum speed of the block and putty:
$m_2v_2 = m_1v_1$
$v_2 = \frac{m_1v_1}{m_2}$
$v_2 = \frac{(2.43~kg)}{2.43~kg+0.50~kg} v_1$
$v_2 = 0.829~v_1$
We can find the original maximum value of kinetic energy:
$K_1 = \frac{1}{2}m_1 v_1^2$
$K_1 = \frac{1}{2}(2.43~kg)(v_1)^2$
$K_1 = 1.215~v_1^2$
We can find the new maximum value of kinetic energy:
$K_2 = \frac{1}{2}m_2 v_2^2$
$K_2 = \frac{1}{2}(2.43~kg+0.50~kg)(0.829~v_1)^2$
$K_2 = 1.0068~v_1^2$
We can find the ratio of $\frac{K_2}{K_1}$
$\frac{K_2}{K_1} = \frac{1.0068~v_1^2}{1.215~v_1^2} = 0.8286$
We can find the new amplitude:
$U_2 = 0.8286~U_1$
$\frac{1}{2}kx_2^2 = 0.8286~\frac{1}{2}kx_1^2$
$x_2^2 = 0.8286~x_1^2$
$x_2 = \sqrt{0.8286}x_1$
$x_2 = (\sqrt{0.8286})(0.20~m)$
$x_2 = 0.18~m$
The new amplitude is $~~0.18~m$
