The period of the glider is 2.6923 s, frequency is 0.3714 Hz, amplitude is 24.0 cm, and maximum speed is approximately 0.5605 m/s.
Step-by-step explanation:
An air-track glider attached to a spring oscillating between the 13.0 cm mark and the 61.0 cm mark completes 13.0 oscillations in 35.0 seconds. We can find the following:
(a) Period: The period is the time it takes to complete one oscillation. It can be calculated by dividing the total time by the number of oscillations. T = 35.0 s / 13.0 = 2.6923 s.
(b) Frequency: The frequency is the number of oscillations per second, which is the inverse of the period. f = 1/T = 1/2.6923 s ≈ 0.3714 Hz.
(c) Amplitude: The amplitude is half the distance between the extreme points of motion. A = (61.0 cm - 13.0 cm) / 2 = 24.0 cm.
(d) Maximum speed: The maximum speed occurs as the glider passes through the equilibrium position. It can be calculated using the formula vmax = 2πfA. vmax = 2π(0.3714 Hz)(0.24 m) ≈ 0.5605 m/s.
The probable question can be: An air-track glider attached to a spring oscillates between the 13.0 {\rm cm} mark and the 61.0 {\rm cm} mark on the track. The glider completes 13.0 oscillations in 35.0 {\rm s}. What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?