Answer
So that the top block does not slip, the maximum speed of the blocks is $0.88~m/s$
Work Step by Step
Let $m_t$ be the mass of the top block. When the blocks accelerate, the force of static friction provides the force to accelerate the 1.0-kg block. We can find the maximum acceleration of the two blocks so that the top block does not slip:
$m_t~g~\mu_s = m_t~a_m$
$a_m = g~\mu_s$
$a_m = (9.80~m/s^2)(0.45)$
$a_m = 4.41~m/s^2$
We can find the angular frequency of the system:
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{\frac{150~N/m}{6.0~kg}}$
$\omega = 5.0~rad/s$
We can find the amplitude of the system:
$a_m = A~\omega^2$
$A = \frac{a_m}{\omega^2}$
$A = \frac{4.41~m/s^2}{(5.0~rad/s)^2}$
$A = 0.176~m$
We can find the maximum speed of the blocks:
$v_m = A~\omega$
$v_m = (0.176~m)(5.0~rad/s)$
$v_m = 0.88~m/s$
So that the top block does not slip, the maximum speed of the blocks is $0.88~m/s$
