In simple harmonic motion (SHM), the maximum speed of a particle occurs when it passes through the equilibrium position. The relationship between the period (T) and the amplitude (A) of SHM can be used to determine the maximum speed.
The period of SHM (T) is the time taken to complete one full oscillation. In this case, T = 0.50 s.
The amplitude (A) is the maximum displacement from the equilibrium position. In this case, A = 20 mm.
The maximum speed (v_max) of the particle in SHM is given by the formula:
v_max = 2πA/T
Substituting the given values, we get:
v_max = 2π * 20 mm / 0.50 s
To calculate the maximum speed, we need to ensure that the units are consistent. Let's convert millimeters (mm) to meters (m) since the SI unit of speed is meters per second (m/s).
1 mm = 0.001 m
v_max = 2π * 0.02 m / 0.50 s
Simplifying the equation, we have:
v_max = 2π * 0.04 m/s
Calculating the value, we find:
v_max ≈ 0.251 m/s
Therefore, the maximum speed of the particle undergoing SHM with a period of 0.50 s and an amplitude of 20 mm is approximately 0.251 m/s.