Final answer:
The angular frequency (ω) of the oscillator can be calculated using the formula ω = 2π/T, where T is the period. The frequency (f) is 1/1.57 Hz. Since the block is released from rest at its maximum displacement, the phase constant θ is 0 radians.
Step-by-step explanation:
To find the angular frequency (ω), we use the given period (T). From the equation T = 2π/ω, we have ω = 2π/T. Given that T is 1.57 seconds, the angular frequency can be calculated as ω = 2π/1.57 s-1. The frequency (f) is the reciprocal of the period, f = 1/T, which simplifies to f = 1/1.57 Hz.
The displacement x(t) for a simple harmonic oscillator is given by x(t) = X cos(ωt + θ), where X is the amplitude and θ is the phase constant. Given that the initial velocity is -92 m/s at t=0 and the amplitude X is 0.02 m, we could equate this to the velocity formula derived from the displacement: v(t) = -ωX sin(ωt + θ). To find the phase constant θ, we set t=0 and solve for θ using the initial conditions.
However, as the block is released from rest at its maximum displacement, the phase constant θ is 0 radians (or multiples of π radians.) Therefore, the complete equation of motion becomes x(t) = 0.02 cos(ωt), and the angular frequency ω can be calculated directly from the period.