Given the problem, we need to determine a function that models the position [tex]\( s(t) \)[/tex] of the weight in terms of time [tex]\( t \)[/tex].
### Problem Breakdown
1. Amplitude and Period:
- The weight is pulled to a maximum displacement of 6 inches from the equilibrium position, so the amplitude is 6 inches.
- The weight completes one full oscillation in 4 seconds, which means the period is 4 seconds.
2. Angular Frequency:
- To find the angular frequency [tex]\( \omega \)[/tex], we use the relationship between the period [tex]\( T \)[/tex] and the angular frequency. The angular frequency is given by:
[tex]\[
\omega = \frac{2\pi}{T}
\][/tex]
Substituting the given period:
[tex]\[
\omega = \frac{2\pi}{4} = \frac{\pi}{2}
\][/tex]
3. Initial Condition:
- Since the weight is pulled down to its maximum displacement initially and then released, it starts at the amplitude position. This means a cosine function that starts at maximum (with a negative amplitude, as it is pulled down) fits well. Thus, a function of the form [tex]\( -A \cos(\omega t) \)[/tex] is appropriate here.
### Determining the Correct Function:
With all elements in place:
- Amplitude [tex]\( A = 6 \)[/tex]
- Angular frequency [tex]\( \omega = \frac{\pi}{2} \)[/tex]
- Since it starts from the maximum displacement (pulled down), the function should have a negative cosine form.
Putting these together, the function becomes:
[tex]\[ s(t) = -6 \cos \left(\frac{\pi}{2} t\right) \][/tex]
Thus, the correct function that models the position of the weight is:
[tex]\[ \boxed{s(t) = -6 \cos \left(\frac{\pi}{2} t\right)} \][/tex]
### Answer:
A. [tex]\( s(t) = -6 \cos \left(\frac{\pi}{2} t\right) \)[/tex]
