Final answer:
In a mass-spring system undergoing simple harmonic motion, when the velocity of the mass is half the maximum value, the displacement from equilibrium can be calculated using the formula x = A / √2. With an amplitude of 16 cm, the displacement where the speed is half the maximum is approximately 11.3 cm.
Step-by-step explanation:
The problem is describing a mass-spring system undergoing simple harmonic motion (SHM). According to SHM theory, the velocity of a mass attached to a spring is proportional to the square root of the spring constant times the mass (v = √(k/m)) and is dependent on the displacement from the equilibrium position. The maximum velocity occurs when the displacement is zero.
For any given instant, the velocity can be described by v = √(k/m)(A2 - x2), where A is the amplitude, x is the current displacement, and k is the spring constant. When the speed is half the maximum, the equation v = √(k/m)(A2 - x2) / 2 must be solved for x.
Setting up the equation with known quantities, we get (√(k/m)(A2 - x2)) / 2 = (√(k/m)A) / 2, which simplifies to x = A / √2. For an amplitude of 16 cm, this gives us a displacement of 16 cm / √2, which is approximately 11.3 cm.