A 4.0-kg block extends a spring 16 cm from its unstretched position. The block is removed
and a 0.50-kg body is hung from the same spring. If the spring is then stretched and released, what is
period of oscillation?
To find the period of oscillation, we need to know the effective spring constant of the spring with the 0.50-kg body hanging from it.
First, let's determine the spring constant of the spring using the data provided when the 4.0-kg block was attached to it.
The force exerted by the spring (F_spring) can be calculated using Hooke's Law:
F_spring = k * x
where k is the spring constant and x is the displacement from the equilibrium position.
Given:
Mass of the 4.0-kg block (m1) = 4.0 kg
Displacement (x) = 16 cm = 0.16 m (converted from cm to m)
The weight of the block is equal to the force exerted by the spring. Thus, we can write:
m1 * g = k * x
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values:
4.0 kg * 9.8 m/s^2 = k * 0.16 m
Simplifying:
k = (4.0 kg * 9.8 m/s^2) / 0.16 m = 245 N/m
Now, let's calculate the effective spring constant when the 0.50-kg body is attached to the spring.
Given:
Mass of the 0.50-kg body (m2) = 0.50 kg
The effective spring constant (k_eff) can be calculated as follows:
k_eff = (m1 * k) / (m1 + m2)
Substituting the given values:
k_eff = (4.0 kg * 245 N/m) / (4.0 kg + 0.50 kg)
Simplifying:
k_eff = 980 N/m
The period of oscillation (T) can be calculated using the formula:
T = 2π * sqrt(m_eff / k_eff)
where m_eff is the effective mass. In this case, m_eff = m1 + m2.
Substituting the given values:
m_eff = 4.0 kg + 0.50 kg = 4.5 kg
T = 2π * sqrt(4.5 kg / 980 N/m) = 2π * sqrt(0.0046 kg/N) ≈ 0.607 seconds
Therefore, the period of oscillation is approximately 0.607 seconds.