Find a model for simple harmonic motion if the position at t = 0 is 4, the amplitude is 4 centimeters, and the period is 6 seconds.

B. d = 4 cos(π/3 t)

Step-by-step explanation:

As at t=0 the amplitude is 4 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at t=0 is 0.  

sin 0=0

We know that,

y = a. cosb(t + c) + d

Period = 2 π / b

Horizontal shift or phase shift is c,

Amplitude is a,

Vertical shift is d.

As in this case the amplitude is given to be 4, hence  = 4

Also the period is given as 6.

That is, 2π/b = 6

b = 2π/6 = π/3

Here horizontal and vertical shifts are 0.

That is c = 0 and d = 0

Plug those values in the above form, we get

y = 4.cos π/3 (t + 0) + 0

y = 4 cos (π/3)t

B. d = 4 cos(π/3 t)

Hope this will helpful.

Thank you.

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1