The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
The maximum speed of the piston is 100 m/min (or approximately 1.67 m/s).
Explanation
This question tests the understanding of the relationship between amplitude, angular frequency, and maximum speed in Simple Harmonic Motion (SHM). From the textbook context, the speed of a particle in SHM is given by v = ωA, where ω is the angular frequency and A is the amplitude. The stroke is defined as twice the amplitude, so amplitude A = stroke/2 = 1.0/2 = 0.5 m. Substituting the given values: maximum speed = 200 rad/min × 0.5 m = 100 m/min.
Solution Steps
Step 1: Given, Stroke = 1.0 m. Since stroke is twice the amplitude, Amplitude (A) = Stroke/2 = 1.0/2 = 0.5 m.
Step 2: Given, Angular frequency (ω) = 200 rad/min.
Step 3: From the context, the speed of a particle in SHM is v = ωA. Therefore, Maximum speed = ω × A.
Step 4: Substituting the values: Maximum speed = 200 rad/min × 0.5 m = 100 m/min.
Step 5: Converting to SI units: 100 m/min = 100/60 m/s = 5/3 m/s 1.67 m/s.