A mass of 0.40 kg,hanging from a spring with a spring constant of 160 N/m,is set into an up-and-down simple harmonic motion.What is the speed of the mass when moving through the equilibrium point? The starting displacement from equilibrium is 0.10 m.
A) zero
B) 1.4 m/s
C) 2.0 m/s
D) 3.4 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we need to use conservation of energy. At the equilibrium point, the mass has purely potential energy and no kinetic energy. As it oscillates, the potential energy is converted into kinetic energy and vice versa. At the farthest point from equilibrium, the mass has purely kinetic energy and no potential energy.

We can use the equation for the total mechanical energy of the system:

E = 1/2 kx^2 + 1/2 mv^2

where k is the spring constant, x is the displacement from equilibrium, m is the mass, and v is the speed.

At the equilibrium point, x = 0.10 m and v = 0, so the total mechanical energy is purely potential energy:

E = 1/2 kx^2 = 1/2 (160 N/m)(0.10 m)^2 = 0.80 J

At the farthest point from equilibrium, x = 0 and the potential energy is zero, so the total mechanical energy is purely kinetic energy:

E = 1/2 mv^2

We can solve for v:

v = sqrt(2E/m) = sqrt(2(0.80 J)/(0.40 kg)) = 2.0 m/s

Therefore, the answer is C.

A Mass of 0