To find the mass of the ball, we can use Hooke's law for the restoring force of the spring:
F = k * x
Where:
F is the force applied by the spring,
k is the spring constant,
x is the displacement from the equilibrium position.
In this case, the displacement of the ball from the equilibrium position is 7.50 cm, which is equal to 0.075 m. The force applied by the spring is equal to the weight of the ball, given by:
F = m * g
Where:
m is the mass of the ball,
g is the acceleration due to gravity (approximately 9.8 m/s²).
Setting these two equations equal to each other, we have:
m * g = k * x
Substituting the given values:
m * 9.8 m/s² = 15.8 N/m * 0.075 m
Solving for m, we get:
m = (15.8 N/m * 0.075 m) / 9.8 m/s²
m ≈ 0.121 kg
Therefore, the mass of the ball is approximately 0.121 kg.
Now, to find the maximum speed of the ball, we can use the relationship between the maximum speed (v_max) and the amplitude (A) of the oscillation:
v_max = ω * A
Where:
ω is the angular frequency,
A is the amplitude of the oscillation.
The angular frequency can be calculated using the formula:
ω = 2π / T
Where:
T is the period of the oscillation.
In this case, the period T is given as 22.0 seconds. Therefore:
ω = 2π / 22.0 s
Substituting this value into the expression for v_max, along with the amplitude A = 0.075 m:
v_max = (2π / 22.0 s) * 0.075 m
Calculating this expression gives us:
v_max ≈ 0.429 m/s
Therefore, the maximum speed of the ball is approximately 0.429 m/s.