Answer:

Explanation:
As we know that amplitude of forced oscillation is given as

here we know that natural frequency of the oscillation is given as

here mass of the object is given as



angular frequency of applied force is given as


now we have


Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
The elastic potential energy stored in the car's spring during the process is 3.75 J
<h3>Determination of the spring constant</h3>
From the question given above, the following data were obtained:
K = F/e
K = 15 / 0.5
K = 30 N/m
<h3>Determination of the potential energy</h3>
- Spring constant (K) = 30 N/m
PE = ½Ke²
PE = ½ × 30 × 0.5²
PE = 15 × 0.25
PE = 3.75 J
Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J
Learn more about energy stored in spring:
brainly.com/question/4280346