Find a model for simple harmonic motion if the position at t=0 is 5 centimeters, the amplitude is 5 centimeters, and the period is 4 seconds.A) d=5cos(4t)B) d=5sin(pi/2t)C) d=4sin(5t)D) d=5cos(pi/2t)

Answer:[tex]\boxed{\boxed{D.\ d=5\cos \left(\dfrac{\pi}{2}t\right)}}[/tex]Step-by-step explanation:As at t=0 the amplitude is 5 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at x=0 is 0. [tex]\sin 0=0[/tex]We know that, in the function[tex]y=a\cdot \cos b(x+c)+d[/tex]period is [tex]\dfrac{2\pi}{b}[/tex], horizontal shift or phase shift is c, amplitude is a, vertical shift is d.As in this case the amplitude is given to be 5, hence [tex]a=5[/tex]Also the period is given as 4, so[tex]\Rightarrow \dfrac{2\pi}{b}=4[/tex][tex]\Rightarrow b=\dfrac{2\pi}{4}=\dfrac{\pi}{2}[/tex]As nothing is given about the horizontal and vertical shift, so putting [tex]c=0,d=0[/tex], the function becomes,[tex]y=5\cdot \cos \dfrac{\pi}{2}(x+0)+0[/tex]or [tex]y=5\cos \left(\dfrac{\pi}{2}x\right)[/tex]As in x axis time is taken as t and in y axis distance is taken as d, so the function becomes,[tex]d=5\cos \left(\dfrac{\pi}{2}t\right)[/tex]