Answer:
A = 0.22 m
Step-by-step explanation:
The spring with the block and the pebble forms an oscillatory system, this system is described by the expression
x = A cos (wt + φ)
w = √ (k / m).
The data they give us is the amplitude of the movement (A = 10 cm), the oscillation mass is equal to the block mass plus the mass of the pebble
m = m + M
m = 0.031 + 0.108
m = 0. 139 kg
To find the spring constant let's use Hooke's law
F = k X
The force is the weight of the pebble and the additional elongation is x = 4.9 cm, let's calculate
k = F / x
k = mg / x
k = 0.031 9.8 / 0.049
k = 6.2 N / m
Let's look for angular velocity
w = √ (6.2 / 0.139)
w = 6,670 rad / s
Let's write the oscillation equation with this data
x = 0.10 cos (6,670 t)
For the pebble to remain in contact with the block, the acceleration of the spring system plus block with pebble must be less than the acceleration of gravity in the descending oscillation
Let's look for system acceleration
a = d²x / dt²
dx / dt = - A w sin (wt + Ф)
d²x / dt² = - A w² cos (wt+Ф)
To find the maximum value cos (wt) = ±1
g = A w²
A = g / w²
A = 9.8 / 6.67²
A = 0.22 m
When the amplitude of the oscillation exceeds this value the pebble is delayed with respect to the block