Answer:
The position of the particle in the xy-plane is given by:
x = A cos(ωt)
y = A sin(ωt)
Differentiating the position equation twice with respect to time gives the acceleration equation:
a = -Aω^2cos(ωt) - Aω^2sin(ωt)
The magnitude of the acceleration is:
|a| = sqrt[(Aω^2cos(ωt))^2 + (Aω^2sin(ωt))^2]
Since the maximum value of sin and cos functions is 1, the maximum magnitude of acceleration is:
|a|max = sqrt[(Aω^2)^2 + (Aω^2)^2] = sqrt[2(Aω^2)^2] = sqrt[2]Aω^2
Substituting the given values gives:
|a|max = sqrt[2](1.5 m)(2.0 rad/s)^2 = 6.0 m/s^2
Therefore, the magnitude of the particle's acceleration is 6.0 m/s2, option (e).
Step-by-step explanation: