Question

If the phase angle for a block–spring system in SHM is ϕ and the block's position is given by x = xm cos(ωt + ϕ), what is the ratio of the kinetic energy to the potential energy at time t = 0? (Use any variable or symbol stated above as necessary.)

Answer 1

K.E/P.E = m/k  tan²φ x ω²

Explanation:

The given position of block x = x₀ cos(ωt + φ)

The velocity of block  v = dx/dt = - x₀ sin(ωt + φ) x ω

The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²

The potential energy of spring = 1/2 k x² , where k is the spring constant

Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)

When t = 0

K.E = 1/2 m x₀²sin²φ x ω²

P.E = 1/2 k x₀² cos²φ

Dividing these , we have

K.E/P.E = m/k  tan²φ x ω²