Final answer:
The amplitude of the oscillation of a spring stretched from an unstretched length of 20 cm to 30 cm by a mass is 5.0 cm. Without the spring constant, the frequency cannot be accurately determined. On the moon, the oscillation frequency would decrease due to lower gravity.
Step-by-step explanation:
The amplitude of an oscillation is the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In this scenario, the spring stretched from 20 cm to 30 cm due to the mass attached. The amplitude, therefore, is half the difference of the stretched length and the unstretched length, which is (30 cm - 20 cm) / 2 = 5 cm. Therefore, the amplitude of the oscillation is 5.0 cm, making the correct answer A. 5.0 cm.
For the frequency of the oscillation, we would need to use the formula f = 1/(2π) ∙ √(k/m), where k is the spring constant and m is the mass. However, since the spring constant is not provided in the question, we cannot calculate an exact numeric value based on the information given. Presumably, there is some miscommunication or missing data, and therefore an accurate frequency cannot be determined from the provided question.
On the Moon, the acceleration due to gravity is approximately 1/16 of that on Earth. This would affect the force the mass exerts on the spring, and hence change the frequency of the oscillation since frequency is dependent on gravity when considering a pendulum or a mass on a spring in a gravitational field. Here, the frequency would decrease because the restoring force driving the oscillation would be weaker due to the reduced gravitational pull, making the correct answer A. The frequency would decrease.