Answer:
The maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is approximately 1.309 seconds.
Step-by-step explanation:
The time period (T) of a mass-spring system is given by:
T = 2π√(m/k)
where m is the mass attached to the spring, and k is the spring constant.
Given that the spring is displaced 5 cm below its equilibrium position and travels from the lowest point to the highest point within 0.25 sec. This means that the time period of the system is:
T = 2(0.25) = 0.5 sec
Now, let's assume that the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is t seconds.
So, the time taken for the system to move from the lowest point to 3 cm above the equilibrium position is (t/2) seconds.
According to the given problem, the displacement is 5 cm below the equilibrium position, so the amplitude of oscillation is:
A = (5 + 3) / 2 = 4 cm
Now, using the formula for time period, we get:
T = 2π√(m/k) ---- (1)
We know that the maximum displacement (amplitude) of oscillation, A = 4 cm. This can be expressed in terms of mass and spring constant as:
A = (m * g) / k ---- (2)
where g is the acceleration due to gravity.
Squaring equation (2) and solving for m/k, we get:
(m/k) = (A * k) / g)^2 ---- (3)
Substituting equation (3) into equation (1), we get:
T = 2π√[((A * k) / g)^2] ---- (4)
Simplifying equation (4), we get:
T = 2π * (A / g) * √(1/k) ---- (5)
Now, substituting the values of T, A, and g into equation (5), we get:
0.5 = 2π * (4 / 9.8) * √(1/k)
Simplifying this equation, we get:
√(k) = 2π * (4 / 9.8) / 0.5
√(k) = 10.239
k = 105
So, the spring constant is 105 N/m.
Now, substituting the value of k into equation (3), we get:
(m/k) = (A * k / g)^2
(m/k) = (4 * 105 / 9.8)^2
(m/k) = 73.88
So, the mass attached to the spring is:
m = (73.88) * (105)
m = 7757.4 g
m = 7.7574 kg
Now, we know the mass of the system and the spring constant, we can calculate the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position.
The time period (T) of the system is given by:
T = 2π√(m/k)
T = 2π√(7.7574/105)
T = 1.309 sec (approx)
Therefore, the maximum time required for the system to oscillate from 5 cm below the equilibrium position to 3 cm above the equilibrium position is approximately 1.309 seconds.