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For the simple harmonic motion equation

d=8sin(4πt), what is the period?

a. 1/2
b. 1/4
c. 1/8
d.1/ π


User Dlopatin
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1 Answer

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Final answer:

The period of the simple harmonic motion equation d=8sin(4πt) can be found by determining the angular frequency and then calculating the reciprocal of the frequency. With an angular frequency of 4π, the frequency is 2, leading to a period of 1/2. Thus, the correct answer is a. 1/2.

Step-by-step explanation:

To find the period of the simple harmonic motion described by the equation d=8sin(4πt), we need to consider the coefficient of t inside the sine function. The general form of a simple harmonic motion equation is x = A sin(ωt), where A is the amplitude, and ω is the angular frequency, which is related to the frequency f by the formula ω = 2πf. The period T is the reciprocal of the frequency, given by the formula T = 1/f.

In the equation provided, the angular frequency ω is 4π. Thus, the frequency f is given by f = ω/(2π), which simplifies to f = 2. Consequently, the period T is the reciprocal of the frequency, giving us T = 1/2. Hence, the correct answer to the period of the simple harmonic motion equation d=8sin(4πt) is a. 1/2.

User Alwinlin
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