A spring with a force constant of 111 N/m is attached to the ceiling. When a mass of 3.78 kg is attached it pulls the spring down .340 meters to its equilibrium position. The mass is then pulled 9cm below its equilibrium position and oscillates back and forth. Determine the period of oscillation (in seconds).
To determine the period of oscillation, we can use the formula:
T = 2π√(m/k)
where T is the period of oscillation, m is the mass (3.78 kg), and k is the force constant (111 N/m).
First, let's calculate the effective mass (m_eff) of the system. When the mass is pulled below the equilibrium position, it stretches the spring and increases the effective mass of the system. The effective mass can be calculated using Hooke's Law:
F = kx
where F is the force exerted by the spring, k is the force constant, and x is the displacement from the equilibrium position. Rearranging the equation, we have:
m_eff = F/k
Given that the mass is pulled 9 cm below its equilibrium position, the displacement x is 0.09 m. Substituting the values into the equation, we get:
m_eff = kx
= (111 N/m)(0.09 m)
= 9.99 kg
The effective mass of the system is approximately 9.99 kg.
Now we can use the formula for the period of oscillation:
T = 2π√(m_eff/k)
Substituting the values, we get:
T = 2π√(9.99 kg / 111 N/m)
= 2π√(0.0900 s^2)
= 2π(0.3 s)
≈ 1.88 s
The period of oscillation is approximately 1.88 seconds.