Final answer:
The period of the simple harmonic motion equation d=8sin(4πt) is T=1/2, since the angular frequency ω is 4π, which gives a frequency f of 2, making the period T the reciprocal of the frequency. Option 1 is correct.
Step-by-step explanation:
The student is asking what the period of the simple harmonic motion equation d=8sin(4πt) is. In this equation, the factor multiplying t inside the sine function determines the frequency of the motion. For a sinusoidal function of time such as Asin(ωt) or Acos(ωt), the angular frequency ω is related to the frequency f by ω = 2πf, and the period T is the reciprocal of the frequency, i.e., T=1/f.
In the given equation, the angular frequency ω is 4π. Therefore, the frequency is f = ω/(2π) = 4π/(2π) = 2. The period T is the reciprocal of the frequency, T=1/f = 1/2. Hence, Option 1: T=1/2 is the correct answer.
To find the period of the simple harmonic motion equation d=8sin(4πt), we can use the formula T = 2π/w, where T is the period and w is the angular frequency. In this case, the angular frequency is 4π, so plugging it into the formula, we get T = 2π/(4π) = 1/2. Therefore, the correct option is Option 1: T=1/2.