Final answer:
When a small ball of mass 'm' is released at a height 'r' above the surface of the Earth and reaches a maximum depth of r/2 inside the Earth through a narrow groove containing an ideal spring, the spring constant 'k' can be calculated using Hooke's law. The correct value of 'k' is 3gmmr³.
Step-by-step explanation:
To solve this problem, we can apply the concept of simple harmonic motion. When a small ball of mass 'm' is released at a height 'r' above the surface of the Earth, it will fall inside the Earth and reach a maximum depth of r/2. Within the narrow groove, the ball will experience an restoring force from the ideal spring.
The force exerted by the spring can be given by Hooke's law: F = -kx, where F is the force, k is the spring constant, and x is the displacement from the natural length of the spring. As the ball reaches the maximum depth, the displacement x will be r/2.
At maximum depth, the force of gravity pulling the ball downward will be balanced by the restoring force from the spring. Therefore, we can equate the two forces and solve for the value of the spring constant k:
mg = -k(r/2)
k = -2mg/r
Substituting the value of g = GM/r^2, where G is the gravitational constant and M is the mass of Earth, we get:
k = -2mGM/r³
Since we want the value of k in terms of 'gmmr³' (where g = GM/r²), we can divide both sides by 'g' to obtain:
k = -2mmr
Therefore, the correct option is (a) 3gmmr³.