Question

Word problem involving a sine or cosine function: Problem type 1

Answer 1

Given:

a.) Amplitude = 9 cm

b.) Period = 0.5 seconds

c.) At time = 0, displacement from rest is -9 cm

The general equation for displacement as a function of time is:

`\text{ d = -A }\cdot\text{ Cos}(\text{Bt - C) }+\text{ D}`

The amplitude is given by A. Here, since amplitude is 9 cm, so we have:

A = 9

The period of the object is 0.5 seconds, so we have:

`\text{ Period = 2}\frac{\text{ }\pi}{\text{ B}}`
`\begin{gathered} 0.5\text{ }=(1)/(2)\text{= 2}\frac{\pi}{\text{ B}} \\ \text{ B = }(\text{ 2}\frac{\pi}{\text{ B}})(2) \\ \text{ B = 4}\pi \end{gathered}`

Since the minimum displacement is at t = 0, so there is no phase shift.

Therefore, C = 0

Also since there is no vertical shift, so D = 0.

In Summary, we get: A = 9, B = 4π, C = 0 and D = 0

Thus the equation modeling the displacement d as a function of time t is given by:

`\text{ d = -A }\cdot\text{ Cos}(\text{Bt - C) }+\text{ D}`
`\text{ d = -9 }\cdot\text{ Cos}(4\pi\text{t - 0) }+0`
`\text{ d = -9Cos}(4\pi\text{t)}`

Therefore, the equation is:

`\text{ d = -9Cos}(4\pi\text{t)}`