Final answer:
The amplitude of the oscillator with the acceleration equation ax(t) = -(22 m/s^2)cos(36t) is approximately 1.7 cm.
Step-by-step explanation:
The acceleration of an oscillator undergoing simple harmonic motion (SHM) is given in the form a(t) = -ω2X cos(ωt), where ω is the angular frequency and X is the amplitude of the oscillator. Given the acceleration equation ax(t)= -(22 m/s2)cos(36t), we can see that the amplitude of acceleration is 22 m/s2. To find the amplitude of the oscillator, we can use the relation for acceleration in SHM:
a(t) = -(ω2)X.
Comparing this with the given acceleration equation, we conclude that ω2X = 22 m/s2. Since ω is given as 36 s-1 (the coefficient of t in the cosine function), ω2 = 362 s-2. Substituting ω2 into the equation, we find that the amplitude X of the oscillator is 22 m/s2 / 362 s-2, which simplifies to approximately 0.017 m or 1.7 cm.