Answer:
x = ± √(8/9) A.
Step-by-step explanation:
For a simple harmonic oscillator with amplitude A and angular frequency ω, the speed at position x is given by v = ω√(A^2 - x^2).
The maximum speed occurs at the equilibrium position, where x = 0, and is given by vmax = ωA.
To determine the positions at which the speed is one-third of the maximum speed, we need to solve the equation:
v = (1/3)vmax
ω√(A^2 - x^2) = (1/3)ωA
√(A^2 - x^2) = (1/3)A
A^2 - x^2 = (1/9)A^2
8/9 A^2 = x^2
x = ± √(8/9) A
Therefore, the positions at which the speed of the simple harmonic oscillator is one-third of the maximum speed are x = ± √(8/9) A.