A2.0-kg mass is oscillating about the origin at 24 rad/s. the amplitude of the oscillations is 0.040 m. at what position is the potential energy equal to twice the kinetic energy? show all work and steps, typed .

0.0327 m

Explanation:

m = 2 kg

ω = 24 rad/s

A = 0.040 m

Let at position y, the potential energy is twice the kinetic energy.

The potential energy is given by

U = 1/2 m x ω² x y²

The kinetic energy is given by

K = 1/2 m x ω² x (A² - y²)

Equate both the energies as according to the question

1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)

y² = 2 A² - 2 y²

3y² = 2A²

y² = 2/3 A²

y = 0.82 A = 0.82 x 0.040 = 0.0327 m

answer: he got 23

explanation:

need to see circuit to get time constant