When a 0.30-kg mass is attached to a vertical spring, the spring stretches by 15 cm. How much mass (total) must be attached to the spring to result in a 0.86-s period of oscillation?
To solve this problem, we need to use Hooke's Law and the equation for the period of oscillation.
First, let's determine the spring constant (k) using Hooke's Law:
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:
F = -kx
Where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring stretches by 15 cm, which is equal to 0.15 m. The mass attached to the spring is 0.30 kg.
Since the stretching of the spring is in the downward direction, the force exerted by the spring is equal to the weight of the mass attached:
F = mg
Where m is the mass attached and g is the acceleration due to gravity (approximately 9.8 m/s²).
Therefore, we can equate these two equations:
mg = kx
Solving for k:
k = mg / x
Substituting the given values:
k = (0.30 kg)(9.8 m/s²) / 0.15 m ≈ 19.6 N/m
Now that we have the spring constant (k), let's calculate the mass required for a specific period of oscillation.
The period (T) of an oscillating mass-spring system can be calculated using the formula:
T = 2π * √(m / k)
Where m is the mass attached to the spring and k is the spring constant.
Rearranging the equation to solve for m:
T² = (4π² / k) * m
Substituting the given period (T = 0.86 s) and the spring constant (k = 19.6 N/m):
(0.86 s)² = (4π² / 19.6 N/m) * m
Simplifying the equation:
0.7396 s² = 0.636 s²/kg * m
Solving for m:
m = 0.7396 s² / (0.636 s²/kg) ≈ 1.159 kg
Therefore, to result in a period of 0.86 seconds of oscillation, a total mass of approximately 1.159 kg must be attached to the spring.