Final answer:
The total expansion of the springs depends on the weight of the massive spring compared to the attached weight of the massless spring. If their weights are equal, then both springs will experience the same expansion. Otherwise, the expansions may not be equal.
Step-by-step explanation:
The question asks whether two springs with equal length and spring constant, one with mass m0 and the other massless with a weight m0/3 attached, will experience equal total expansion. This question pertains to the concepts of Hooke's law and the effect of masses on spring expansion. In compliance with Hooke's Law, spring extension is directly proportional to the force applied to the spring up to the elastic limit of the spring. We are considering two scenarios: a massive spring with its weight and a massless spring with an attached weight.
In physics, especially when discussing simple harmonic motion (SHM) and springs, Hooke's Law determines that the force F exerted by a spring is equal to the negative product of the spring constant k and the displacement x from the spring's original length, or F = -kx. When a mass m0 is hung from a spring, the gravitational force (weight) of the mass causes the spring to stretch until the spring's restoring force balances the weight. Spring constant, represented as k, is a measure of a spring's resistance to being compressed or stretched
If the mass of the first spring is included in its analysis (since it is massive), it will stretch due to both its weight and the force exerted by any additional external load. On the other hand, the massless spring will stretch due only to the external weight attached. Therefore, if the mass of the first spring is such that its weight alone causes it to stretch as much as the mass attached to the second spring, both springs would experience the same expansion. However, the problem does not clearly state the weight of the massive spring compared to the attached weight of the massless spring. The comparison of expansions cannot be determined definitively without knowing the weight of the massive spring or assuming it equals to the weight added on the massless spring.