Final answer:
The period of oscillation for a 442 g mass on a spring with a spring constant of 7.31 N/m is calculated using the formula T = 2π√(m/k). After converting the mass to kilograms and substituting into the formula, the period of oscillation is found to be approximately 3.115 seconds.
Step-by-step explanation:
To determine the period of oscillation for a mass attached to a spring undergoing simple harmonic motion (SHM), we can use the formula:
T = 2π√(m/k)
Where:
- T is the period of oscillation in seconds.
- m is the mass in kilograms (kg).
- k is the spring constant in newtons per meter (N/m).
Given the mass of 442 g, which is equivalent to 0.442 kg, and a spring constant k = 7.31 N/m, we can calculate the period T as follows:
T = 2π√(0.442 kg / 7.31 N/m)
Now, performing the calculation:
T = 2π√(0.0604 kg⋅m/N)
T ≈ 2π√(0.0604) ≈ 2π√(0.2458)
T ≈ 2π(0.4958)
T ≈ 3.115 seconds
Therefore, the period of oscillation for a 442 g mass on a spring with a constant k = 7.31 N/m is approximately 3.115 seconds.