Position of particle:

x = A sin(ωt)

Potential energy is minimum at:

t = T / 2β

Find minimum positive value of β.

Solution:

Potential energy:

U = (1/2)kx²

Potential energy is minimum when:

x = 0

For:

x = A sin(ωt)

sin(ωt)=0

ωt = nπ

Smallest positive time:

t = π/ω

Since:

ω = 2π/T

t = T/2

Given:

T/(2β) = T/2

β = 1

Answer: 1