Final answer:
For two springs with different spring constants but attached to identical masses and stretched the same distance from equilibrium, the spring with the lower spring constant will have a longer period of oscillation. The period of oscillation is inversely proportional to the square root of the spring constant.
Step-by-step explanation:
The question seeks to understand how the period of oscillation would change between two different springs when they are attached to identical masses and stretched the same distance from their equilibrium positions. Given that one spring has a spring constant (k) twice as much as the other, we need to evaluate the effect of the spring constant on the period of oscillation. The period of oscillation for a mass-spring system is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.
When the spring constant is doubled, as is the case for the first spring in comparison to the second spring, it would cause the period of oscillation to be shorter, given that the spring constant k appears in the denominator of the equation and the masses are identical. To make it simpler, if spring A has a spring constant 2k and spring B has a spring constant k, we expect that the period for spring A would be shorter than the period for spring B when the same mass is attached. Therefore, the spring with the lower spring constant (the second spring) will have the longer period of oscillation.