Final answer:
The speed of a simple harmonic oscillator is one third of its maximum speed at positions approximately ±3.77 cm from the equilibrium position when the amplitude is 4.00 cm.
Step-by-step explanation:
To find the positions where the speed of a simple harmonic oscillator is one third of its maximum speed, we will use the relationship between velocity and position in simple harmonic motion (SHM). The maximum speed Vmax occurs at the equilibrium position (x=0) and is given by Vmax = ωA, where ω is the angular frequency, and A is the amplitude.
Using the conservation of energy for SHM, the kinetic energy at any position x is equal to the difference between the maximum potential energy and the potential energy at position x. This gives:
K.E. = ½ mω²(A²-x²)
Since kinetic energy K.E. is proportional to the square of velocity, we can write:
(v/Vmax)² = (A²-x²)/A²
For a speed of v = Vmax/3, we square both sides of this equation:
(1/3)² = (A²-x²)/A²
Solving for x gives:
x² = A² - (A²/9)
x² = (8/9)A²
x = ±(√(8/9))A
Substituting the given amplitude A = 4.00 cm:
x = ±(√(8/9))(4.00 cm)
x ≈ ±3.77 cm
Therefore, the particle's speed equals one third of its maximum speed at approximately ±3.77 cm from the equilibrium position.