Test

4. An object's motion is described by the equation [tex] d = 3 \sin (B=n-2) [/tex]. The displacement, [tex] d [/tex], is in meters, and the time, [tex] t [/tex], is in seconds. Answer the following questions.

Part I: What is the object's position (in meters) at [tex] t=0 [/tex]? (4 points)
[tex]\[ d = 3 \sin (0) - 2 \][/tex]
[tex]\[ d = 0 - 2 \][/tex]
[tex]\[ d = -2 \][/tex]

Part II: What is the object's maximum displacement (in meters) from its [tex] t=0 [/tex] position?

Part III: How much time (in seconds) is required for one oscillation? (2 points)



Answer :

Alright, let's tackle this problem step-by-step.

### Part I: Object's position at [tex]\( t = 0 \)[/tex]

The equation for the object's displacement is given by:

[tex]\[ d = 3 \sin(t) - 2 \][/tex]

To find the position at [tex]\( t = 0 \)[/tex], substitute [tex]\( t = 0 \)[/tex] into the equation:

[tex]\[ d = 3 \sin(0) - 2 \][/tex]

Since [tex]\(\sin(0) = 0\)[/tex]:

[tex]\[ d = 3 \cdot 0 - 2 \][/tex]
[tex]\[ d = -2 \][/tex]

Thus, the object's position at [tex]\( t = 0 \)[/tex] is:

[tex]\[ \boxed{-2 \text{ meters}} \][/tex]

### Part II: Maximum displacement from the [tex]\( t = 0 \)[/tex] position

The sine function, [tex]\(\sin(t)\)[/tex], oscillates between -1 and 1.

1. Maximum displacement:

For the maximum value of [tex]\(\sin(t) = 1\)[/tex]:

[tex]\[ d_{\text{max}} = 3 \sin(1) - 2 \][/tex]
[tex]\[ d_{\text{max}} = 3 \cdot 1 - 2 \][/tex]
[tex]\[ d_{\text{max}} = 1 \][/tex]

2. Minimum displacement:

For the minimum value of [tex]\(\sin(t) = -1\)[/tex]:

[tex]\[ d_{\text{min}} = 3 \sin(-1) - 2 \][/tex]
[tex]\[ d_{\text{min}} = 3 \cdot (-1) - 2 \][/tex]
[tex]\[ d_{\text{min}} = -5 \][/tex]

Therefore, the maximum and minimum displacements from the [tex]\( t = 0 \)[/tex] position are:

[tex]\[ \boxed{1 \text{ meter}} \ \text{(maximum displacement)} \][/tex]
[tex]\[ \boxed{-5 \text{ meters}} \ \text{(minimum displacement)} \][/tex]

### Part III: Time for one oscillation

The period of the sine function [tex]\(\sin(t)\)[/tex] is [tex]\(2\pi\)[/tex], but since we have [tex]\(\sin(3t)\)[/tex], the period is scaled by the factor inside the sine function. To find the period of [tex]\(3 \sin(t)\)[/tex], we use the formula [tex]\( \text{Period} = \frac{2\pi}{\text{frequency}} \)[/tex]. Here, the frequency is 3:

[tex]\[ \text{Period} = \frac{2\pi}{3} \][/tex]

Thus, the time required for one complete oscillation is:

[tex]\[ \boxed{2.094 \text{ seconds}} \][/tex]