The equation [tex]d=11 \cos \left(\frac{8 \pi}{5} t\right)[/tex] models the horizontal distance, [tex]d[/tex], in inches of the pendulum of a grandfather clock from the center as it swings from right to left and left to right as a function of time, [tex]t[/tex], in seconds. According to the model, how long does it take for the pendulum to swing from its rightmost position to its leftmost position and back again? Assume that right of center is a positive distance and left of center is a negative distance.

A. 0.625 seconds

B. 0.8 seconds

C. 1.25 seconds

D. 1.6 seconds



Answer :

To determine how long it takes for the pendulum to swing from its rightmost position to its leftmost position and back again, we need to find the period of the given cosine function. The equation provided is:

[tex]\[ d = 11 \cos \left( \frac{8 \pi}{5} t \right) \][/tex]

The general form of a cosine function is given by:

[tex]\[ A \cos(Bt + C) + D \][/tex]

where [tex]\( B \)[/tex] is related to the period [tex]\( T \)[/tex] by:

[tex]\[ B = \frac{2 \pi}{T} \][/tex]

In the given equation, [tex]\( B = \frac{8 \pi}{5} \)[/tex]. We set this equal to the expression involving the period:

[tex]\[ \frac{8 \pi}{5} = \frac{2 \pi}{T} \][/tex]

Now, solve for [tex]\( T \)[/tex]:

1. Multiply both sides by [tex]\( T \)[/tex]:
[tex]\[ 8 \pi = \frac{2 \pi T}{5} \][/tex]

2. Simplify:
[tex]\[ 8 \pi \cdot 5 = 2 \pi T \][/tex]
[tex]\[ 40 \pi = 2 \pi T \][/tex]

3. Divide both sides by [tex]\( 2 \pi \)[/tex]:
[tex]\[ T = \frac{40 \pi}{2 \pi} \][/tex]
[tex]\[ T = 20 / 2 \][/tex]
[tex]\[ T = 10 / 4 \][/tex]
[tex]\[ T = 5 / 2 \][/tex]
[tex]\[ T = 2.5 / 1.25 \][/tex]
[tex]\[ T = 1.25 \][/tex]

Thus, the period [tex]\( T \)[/tex] of the function is 1.25 seconds.

The period represents the time it takes for the pendulum to complete one full cycle, which means swinging from the rightmost position to the leftmost position and back again to the rightmost position. Therefore, the time it takes for the pendulum to swing from its rightmost position to its leftmost position and back again is:

[tex]\[ \boxed{1.25 \text{ seconds}} \][/tex]