A massless spring, with K=425 N/m, is hung vertically with the bottom end of the spring at position Y.i. A 3.50 kg mass is attached to the bottom of the spring and allowed to fall and oscillate.
A) How far below Y.i does the mass fall before coming momentarily to rest and reversing direction?
B) What is the amplitude of the oscillations?
C) What is the period of the oscillations?
To solve this problem, we need to use the principles of Hooke's law and simple harmonic motion.
A) To determine how far below Y.i the mass falls before coming to rest and reversing direction, we need to find the equilibrium position of the system. In this case, the equilibrium position is at Y.i. Given that the mass is initially at rest at Y.i, and the spring is initially unstrained, we can set up an equation using Hooke's law:
F = -kx
Where F is the force exerted by the spring, k is the spring constant (425 N/m), and x is the displacement from the equilibrium position.
Since the mass is at rest, the net force on it is zero. At the equilibrium position, the spring force is balanced by the weight of the mass:
mg = kΔx
Where m is the mass (3.50 kg), g is the acceleration due to gravity (9.8 m/s^2), and Δx is the displacement from the equilibrium position.
Solving for Δx:
Δx = mg/k
= (3.50 kg) * (9.8 m/s^2) / (425 N/m)
≈ 0.0818 m
Therefore, the mass falls approximately 0.0818 m below Y.i before coming momentarily to rest and reversing direction.
B) The amplitude of the oscillations can be calculated using the total energy of the system. In simple harmonic motion, the total mechanical energy is given by:
E = (1/2)kA^2
Where E is the total mechanical energy and A is the amplitude of the oscillations.
At the equilibrium position, all the potential energy is converted to kinetic energy, therefore:
E = (1/2)kΔx^2
Plugging in the values:
E = (1/2)(425 N/m)(0.0818 m)^2
≈ 1.420 J
Since the total mechanical energy remains constant throughout the motion, we can set this equal to the potential energy at the extreme position:
E = (1/2)kA^2
Plugging in the values:
1.420 J = (1/2)(425 N/m)A^2
Solving for A:
A^2 = (2 * 1.420 J) / (425 N/m)
A ≈ 0.23 m
Therefore, the amplitude of the oscillations is approximately 0.23 m.
C) The period of the oscillations can be calculated using the equation:
T = 2π√(m/k)
Plugging in the values:
T = 2π√((3.50 kg)/(425 N/m))
≈ 1.71 s
Therefore, the period of the oscillations is approximately 1.71 seconds.