The period of motion of the two blocks when d is small enough to have no slipping is approximately 1.225 s and the coefficient of static friction μ between the surfaces of the two blocks is 0.47 (approximately).
The period of the motion for a mass-spring system can be calculated using the formula:
![\[ T = 2\pi\sqrt{(m)/(k)} \]](https://img.qammunity.org/2024/formulas/physics/high-school/ovy0w3fikmutvs43nv4i50ljas5yvyvv9d.png)
However, in the case of two blocks connected by a spring with static friction, the system will have two normal modes of oscillation. The general expression for the period of such a system is given by:
![\[ T = 2\pi\sqrt{(m)/(k)} \sqrt{(m+M)/(m)} \]](https://img.qammunity.org/2024/formulas/physics/high-school/dcytpfb69y6h4n85a8c0q6jjr8rg16d1f2.png)
Substituting the given values:
=
(0.045)(4.36) = 1.225 s (approximately).
To find the coefficient of static friction between the surfaces of the two blocks,
Given:
The force constant of the spring, k = 250 N/m.
d = 8.8 cm = 0.088 m
The maximum force of static friction can be calculated using the equation:

where
is the normal force between the blocks, and
is the coefficient of static friction.
Since there is no slipping, the normal force is equal to the weight of the top block:
= m * g
where m is the mass of the top block and g is the acceleration due to gravity.
= 0.500 kg * 9.8 m

The spring constant can be related to the maximum force of static friction using the equation:

Combining the equations,
Substituting the given values into the equation, we get:
= 0.47
The coefficient of static friction is approximately 0.47 (rounded to two significant figures).