Final answer:
The period of the glider's oscillation is 2.067 s, the frequency is 0.484 Hz, the amplitude is 23.5 cm, and the maximum speed is 0.722 m/s.
Step-by-step explanation:
To find the period of oscillation, we can divide the total time taken by the number of oscillations. In this case, the total time is 31.0 s and the number of oscillations is 15. So the period (T) of the glider is 31.0 s / 15 = 2.067 s.
The frequency (f) of oscillation can be found by taking the reciprocal of the period, that is, f = 1 / T. So the frequency of the glider is 1 / 2.067 s = 0.484 Hz.
The amplitude of the glider's oscillation is the maximum displacement from its equilibrium position. In this case, the glider oscillates between the 10.0 cm mark and the 57.0 cm mark, so the amplitude (A) is half the difference between these two values. A = (57.0 cm - 10.0 cm) / 2 = 23.5 cm.
The maximum speed of the glider can be found using the formula vmax = ωA, where vmax is the maximum speed, ω is the angular frequency (equal to 2πf), and A is the amplitude. Plugging in the values, we get vmax = 2π(0.484 Hz)(0.235 m) = 0.722 m/s.