Question

A buoy floating in the ocean is bobbing in simple harmonic motion with period 5 seconds and amplitude 4ft. Its displacement d from sea level at time =t0 seconds is −4ft, and initially it moves upward. (Note that upward is the positive direction.) Give the equation modeling the displacement d as a function of time t.

Answer 1

Answer:

$d=-4Cos(\frac{2\pi}{5}t)$

Step-by-step explanation:

We are given that

Period =5 s

Amplitude=4 ft

Displacement d from sea level at time $t=0s=-4 ft$

We have to find the modelling equation displacement d as a function of time.

We know that

The general equation of sinusoidal function is given by

$y(t)=Acos(Bt-C)+D$

B=$\frac{2\pi}{period}=\frac{2\pi}{5}$

When t=0, y=d=-4 ft, D=0

Substitute the values then we get

$-4=4Cos(\frac{2\pi}{5}(0)-C)+0$

$-4=4Cos(-C)$

$Cos(-C)=-1$

We know that Cos(-x)=Cos x

$Cos C=-1$

$Cos C=Cos \pi$  ($cos(\pi)=-1$)

$C=\pi$

Substitute the values then, we get

$d=4Cos(\frac{2\pi}{5}t-\pi)+0$

$d=4Cos(-(\pi-\frac{2\pi}{5}t))$

$d=4Cos(\pi-\frac{2\pi}{5}t)$

$Cos(\pi-x)=-Cosx$

$d=-4Cos(\frac{2\pi}{5}t)$