Write an Equation

Question

An object suspended from a spring is oscillating up and down. The distance from the high point to the low
point is 45 cm, and the object takes 4 s to complete 5 cycles. For the first few cycles, the motion is
modelled by a sine function, where the distance from the mean position, d(t), is measured in centimetres,
with respect to time, in seconds. Write an equation that describes the distance of the object from its mean position

Answer 1

Let t = number of seconds

      d(t) =signed distance from equilibrium position (mean                                    position)at time t  (d(t)>0 when the object is above                            equilibrium , d(t)<0 when the object is below                                        equilibrium, and d(t) = 0 if the object is at the equilibrium                    position).  

 

d(t) = Asin(Bt) assuming that the object is at the equilibrium position at time t = 0.  

 

lAl = maximum distance from equilibrium that the object achieves = 45/2 = 22.5

 

Since the object takes 4 seconds to complete 5 cycles, the period is 4/5 of a second.  

 

The period of a function of the form d(t) = Asin(Bt) (B>0) is 2π/B.

 

          So, 2π/B = 4/5

 

          Cross multiply to get 4B = 10π

 

                                                 B = 5π/2

 

Therefore, d(t) = ±22.5sin(5πt/2)

 

Note:  The values of d(t) range from -22.5 to 22.5.  For the function d(t) = 22.5sin(5πt/2), the object starts at the equilibrium position at time 0 and then starts moving upward. On the other hand, if the object first moves downward, then the function would be             d(t) = -22.5sin(5πt/2).