Question

A piston in a gasoline engine is in simple harmonic motion. The engine is running at the rate of 3 410 rev/min. Taking the extremes of its position relative to its center point as ±4.50 cm.

(a) Find the magnitude of the maximum velocity of the piston.
m/s

(b) Find the magnitude of the maximum acceleration of the piston
km/s²

Answer 1

The displacement can be written as x = A cos(t) ...(1)
Where A is the amplitude of motion and is the angular velocity.
Differentiating equation (1),
dx/dt = velocity = - A sin(t) ...(2)
Maximum velocity, V_max = A

Differentiating equation (2),
dx²/dt² = acceleration = A² cos(t)
Maximum acceleration, a_max = A²

Angular velocity, = 410 rev/min
= (410 x 2) rad / 60 s
= 42.94 rad/s
Amplitude, A = 0.045 m

a)
Maximum velocity, V_max = A
= 0.045 x 42.94
= 1.93 m/s

b)
Maximum acceleration, a_max = A²
= 0.045 x 42.94²
= 82.95 m/s²
= 0.08295 km/s²