Final answer:
a. The maximum force acting on the body of mass 0.2kg is executing SHM with an amplitude force of 20 millimeters if it is 0.064 N is 3.2 N/m.
b. The maximum velocity of the body is 0.0452 m/s.
c. The period of oscillation is 2.78 s.
Step-by-step explanation:
To calculate the maximum force acting on the body, we can use the formula F = -kA, where F is the maximum force, k is the force constant, and A is the amplitude. Rearranging the formula, we get k = -F/A. Substituting the given values, we have
k = -0.064 N / 0.02 m = -3.2 N/m
The maximum velocity can be calculated using the formula v = Aω, where v is the maximum velocity and ω is the angular frequency. The angular frequency, in turn, can be calculated using the formula ω = √(k/m), where m is the mass of the body. Substituting the given values, we have
ω = √(-3.2 N/m / 0.2 kg) ≈ 2.26 rad/s
Substituting this value into the formula for v, we have
v = 0.02 m * 2.26 rad/s = 0.0452 m/s.
The period of oscillation can be calculated using the formula T = 2π/ω, where T is the period and ω is the angular frequency. Substituting the given value of ω, we have
T = 2π / 2.26 rad/s ≈ 2.78 s