Final answer:
To determine when the speed of a simple harmonic oscillator is half its maximum, we solve the velocity equation for when v = ±Vmax/2. The solution reveals that this occurs at positions x = ±X/sqrt(2), which means the object’s speed is half its maximum when it is at approximately 70.7% of its amplitude from the equilibrium position.
Step-by-step explanation:
At what positions is the speed of a simple harmonic oscillator half its maximum? This involves understanding the physics of a mass-spring system. To find the values of position where the speed is half its maximum, we can utilize the relationship between the speed v, maximum speed Vmax, and amplitude X.
For a simple harmonic oscillator, the maximum speed Vmax is achieved at the equilibrium position, which is when x = 0. The velocity as a function of displacement x can be described by the equation v = ±Vmax sqrt(1 - (x/X)^2).
Therefore, to find when the speed is half the maximum speed (v = ±Vmax/2), we set the left side of the equation to Vmax/2 and solve for x/X.
After simplifying, we find that this occurs at positions x = ±X/sqrt(2) relative to the equilibrium position.
Hence, when the spring is displaced by 1/sqrt(2) (approximately 0.707) times the amplitude in either direction from the equilibrium, the speed of the mass will be half the maximum speed.