Question

WILL MARK BRAINLIESTFor the simple harmonic motion equation d =4sin(8pit) what is the period?

Answer 1

The period of any wave is the time it takes for its angle
to go from zero to 2pi .

The 'sin' function is a wave.  The angle of this one is (8pi t).

When t=0, the angle is zero.
Wonderful.
Now, how long does it take for the angle to grow to 2pi ?

I*n other words, when is (8pi t) = 2pi ?

Divide each side by '2pi': . . . . . 4 t = 1

Divide each side by ' 4 ': . . . . . t = 1/4

And there you are.  Every time 't' grows by 1/4, (8pi t) grows by 2pi.
So if you graph this simple harmonic motion described by 'd', you'll
see the graph wiggle up and down with a period of 1/4 . 

Answer 2

Answer: The time period or simply the period of the simple harmonic motion equation d =4sin(8pit) is 1/4.

It is given that  the simple harmonic motion equation d =4sin(8pit).

It is required to find the period or time period.

What is the period of  the simple harmonic motion equation d =4sin(8pit)?

As we know the representation of simple harmonic motion is:

y = A sin ωt

Where A is amplitude and ω is angular frequency.

As we know the time period, t = $\frac{2\pi }{\omega\\}$

From question,

d =4sin(8pit) where ω is 8$\pi$. Putting the value in period,

t = $\frac{2\pi }{\omega\\}=\frac{2\pi }{8\pi }=\frac{1}{4}$

Therefore, the period or time period of the simple harmonic motion equation d =4sin(8pit) is 1/4.

Learn more about simple harmonic motion here:

https://brainly.com/question/14586609

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