A spring streches by 3.90 cm when a 10g mass is hung from it. If a 25g mass attached to this spring oscillates in simple harmonic motion, calculate the period of the motion?
To calculate the period of the motion, we need to use Hooke's Law and the formula for the period of a mass-spring system.
1. First, let's calculate the spring constant, denoted by "k." Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is given by:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium.
In this case, the displacement is 3.90 cm, which we need to convert to meters:
x = 3.90 cm = 0.039 m
The force exerted by the spring is equal to the weight of the mass:
F = m * g
where m is the mass and g is the acceleration due to gravity (9.8 m/s²).
For the 10 g mass:
F = (0.01 kg) * (9.8 m/s²) = 0.098 N
Plugging this into Hooke's Law, we can solve for the spring constant:
k = -F / x = -0.098 N / 0.039 m ≈ -2.51 N/m
2. Now, let's calculate the period of oscillation using the formula:
T = 2π * √(m / k)
where T is the period, m is the mass, and k is the spring constant.
For the 25 g mass:
m = 0.025 kg
Plugging in the values:
T = 2π * √(0.025 kg / (-2.51 N/m))
T ≈ 2π * √(-0.00996 kg/N)
Note that the negative sign appears because the spring force is directed opposite to the displacement.
Simplifying:
T ≈ 2π * √(0.00996 kg/N)
T ≈ 2π * 0.0998 s
T ≈ 0.628 s
Therefore, the period of the motion is approximately 0.628 seconds.