Final answer:
The motion of a 2.4-kg object attached to a spring exhibits simple harmonic motion with a frequency of approximately 10.91 Hz, a period of roughly 0.092 s, an amplitude of 0.1 m, and a maximum speed of about 6.88 m/s.
Step-by-step explanation:
The question asks for the characteristics of simple harmonic motion (SHM) of an object attached to a spring on a frictionless surface when it is released from a stretched position. To find these, we can make use of the formulas related to SHM for a mass-spring system:
- Frequency (f) is calculated using the formula f = (1/2π) * √(k/m), where k is the spring constant and m is the mass.
- Period (T) is the inverse of frequency, so T = 1/f.
- Amplitude (A) is the maximum extent of displacement from the equilibrium, in this case, the amount the spring was stretched initially.
- Maximum speed (Vmax) occurs as the mass passes through the equilibrium point and can be calculated using the formula Vmax = A * 2πf.
For the given 2.4-kg object and a spring constant of 4500 N/m, stretched 0.1 m (10 cm) from equilibrium, we find:
- Frequency: f = (1/2π) * √(4500/2.4) ≈ 10.91 Hz
- Period: T = 1/f ≈ 1/10.91 ≈ 0.092 s
- Amplitude: A = 0.1 m (since the spring is stretched 10 cm from equilibrium)
- Maximum speed: Vmax = A * 2πf ≈ 0.1 m * 2π * 10.91 Hz ≈ 6.88 m/s