A light spring stretches 0.13 m when a 0.35-kg mass is hung from it. The mass is pulled down from this equilibrium position an additional 0.12 m and is released. Determine the spring constant k and the amplitude of vibration.
A) 26.4 N/m and 0.25 m
B) 26.4 N/m and 0.12 m
C) 2.70 N/m and 0.12 m
D) 2.70 N/m and 0.25 m
E) 26.4 N/m and 0.13 m

Question

Quizplus · Accepted Answer

The spring constant can be found using Hooke's Law:

F = kx

where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In this case, the force applied to the spring is the weight of the mass:

F = mg

where g is the acceleration due to gravity (9.8 m/s^2).

So we can write:

mg = kx

Solving for k:

k = mg/x

Substituting the given values:

k = (0.35 kg)(9.8 m/s^2)/(0.13 m)

k = 26.4 N/m

To find the amplitude of vibration, we can use the fact that energy is conserved in a simple harmonic motion system:

PE at max displacement = KE at equilibrium

where PE is potential energy, KE is kinetic energy, and max displacement is the farthest point the mass moves from its equilibrium position.

At the max displacement, all of the potential energy is converted to kinetic energy, so we can write:

1/2 kx^2 = 1/2 mv^2

where v is the velocity of the mass at max displacement.

Using conservation of energy, we know that the maximum velocity occurs at x = 0, so:

v = ±sqrt(k/m) x

Substituting the given values:

v = ±sqrt((26.4 N/m)/(0.35 kg)) (0.12 m)

v = ±0.87 m/s

The amplitude of vibration is equal to the maximum displacement, which is twice the distance from the equilibrium position to the max displacement:

amplitude = 2(0.13 m + 0.12 m)

amplitude = 0.50 m

Therefore, the answer is B: 26.4 N/m and 0.12 m.

A Light Spring Stretches 0