Final answer:
The student's question concerns the calculation of the maximum speed of a mass attached to a spring in simple harmonic motion. The period of the oscillation derived from the given time allows determination of the spring constant, which, along with the initial stretch, is used to calculate the maximum kinetic energy and, subsequently, the maximum speed.
Step-by-step explanation:
The student's question involves finding the maximum speed of a mass attached to a spring and undergoing simple harmonic motion. Given that the mass comes to rest for the first time after 0.744 seconds, we know that this time represents half of a full oscillation period (T/2), as the speed is zero at the endpoints of the oscillation. Thus, the full period (T) is 0.744 s * 2 = 1.488 s.
Using the formula for the period of a simple harmonic oscillator, T = 2Π√(m/k), where m is the mass and k is the spring constant, we can solve for k:
T = 2Π√(0.514 kg/k) => k = (2Π/1.488 s)² * 0.514 kg
With the spring constant k known, we can determine the maximum potential energy (PEmax) of the system when the spring was initially stretched, using PE = (1/2)kx², where x is the stretch (0.0926 m).
Since the potential energy is completely converted into kinetic energy (KE) at the point of maximum speed, KE = (1/2)mv² = PEmax, we can solve for the maximum speed (vmax).