A 1024 g mass is hung on a spring. As a result the spring stretches 35 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? Give your answer in seconds with 3 significant figures.
To find the period of the resulting oscillation, we can use the formula:
T = 2π√(m/k)
Where:
T = Period of oscillation,
m = Mass attached to the spring, and
k = Spring constant.
In this case, we are given the mass of 1024 g (or 1.024 kg) and the stretch of the spring, which is 35 cm (or 0.35 m). To find the spring constant, we can use Hooke's Law:
F = kΔx
Where:
F = Force exerted by the spring,
k = Spring constant, and
Δx = Displacement or stretch of the spring.
In this case, the force exerted by the spring can be calculated using the weight of the mass:
F = mg
Where:
m = Mass, and
g = Acceleration due to gravity (approximately 9.8 m/s²).
Calculating the force:
F = (1.024 kg) * (9.8 m/s²)
F = 10.03 N
Now we can solve for the spring constant:
k = F / Δx
k = 10.03 N / 0.35 m
k ≈ 28.66 N/m
With the spring constant (k) determined, we can now calculate the period (T) using the formula mentioned earlier:
T = 2π√(m/k)
T = 2π√(1.024 kg / 28.66 N/m)
T ≈ 4.167 s
Therefore, the period of the resulting oscillation is approximately 4.167 seconds (s).