A 13.6 kg mass is attached to a spring with a spring constant of 116 N/m. The mass is then stretched to a position 17 m away from the equilibrium position and released. Assuming the system undergoes simple harmonic motion, what is the period of oscillation?

In this case, the period of oscillation for the system with the 13.6 kg mass attached to the spring with a spring constant of 116 N/m, and stretched 17 m away from the equilibrium position, is approximately 2.15 seconds.

1. Given data:

- Mass (m) = 13.6 kg

- Spring constant (k) = 116 N/m

- Displacement from equilibrium position (x) = 17 m

2. The period of oscillation (T) for a mass-spring system can be calculated using the formula:

T = 2π * √(m / k)

3. Substitute the given values into the formula:

T = 2π * √(13.6 / 116)

4. Calculate the period of oscillation:

T = 2π * √(0.1172)

T = 2π * 0.3424

T ≈ 2.15 seconds

Therefore, the period of oscillation for the system with the 13.6 kg mass attached to the spring with a spring constant of 116 N/m, and stretched 17 m away from the equilibrium position, is approximately 2.15 seconds.