A spring is hung from the ceiling. A 0.477 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.153 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

To solve this problem, we can use the concepts of Hooke's Law and simple harmonic motion.

(a) To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this is represented as F = -kx, where F is the force, k is the spring constant, and x is the displacement.

In this case, the block drops a distance of 0.153 m before momentarily coming to rest. At this point, the force exerted by the spring is equal to the weight of the block. Therefore, we can set up the following equation:

kx = mg

Where m is the mass of the block (0.477 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and x is the displacement (0.153 m).

Rearranging the equation to solve for k:
k = mg / x = (0.477 kg)(9.8 m/s^2) / 0.153 m

So the spring constant of the spring is k ≈ 31.18 N/m.

(b) The angular frequency (ω) of the block's vibrations is related to the spring constant (k) and the mass (m) of the block by the formula:

ω = √(k/m)

Substituting the known values:
ω = √(31.18 N/m / 0.477 kg)

Evaluating this expression, we find that the angular frequency of the block's vibrations is approximately ω ≈ 10.29 rad/s.