Answer :
Let's go through the solution step-by-step:
The problem involves a pendulum with a horizontal distance traveled of 12 inches from right to left. This distance corresponds to the amplitude of the motion. Given that the pendulum takes 1 second to swing back from left to right, we understand that the complete period of the pendulum’s motion (one full cycle) is 2 seconds.
Given:
- The amplitude of the motion, which is the maximum distance from the center, is 6 inches. Thus, \( a = 6 \).
1. Determine the period:
The period is the time taken for one complete cycle of the pendulum. Since it takes 1 second to swing back from left to right, the total period (swinging both ways) is 2 seconds.
[tex]\[ \text{Period} = 2 \text{ seconds} \][/tex]
2. Determine the value of \( b \):
The value of \( b \) is derived from the period of the cosine function. The relationship is given by:
[tex]\[ b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{2} = \pi \][/tex]
3. Write the cosine function:
Combining the amplitude and the value of \( b \), the distance \( d \) of the pendulum from the center as a function of time \( t \) is given by:
[tex]\[ d = 6 \cos(\pi t) \][/tex]
4. Find the least positive value of \( t \) at which the pendulum is in the center:
The pendulum is in the center when \( d = 0 \). For the function \( d = 6 \cos(\pi t) \), the value of \( d \) is zero when:
[tex]\[ \pi t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \][/tex]
The least positive \( t \) corresponds to:
[tex]\[ t = \frac{1}{2} = 0.5 \text{ seconds} \][/tex]
5. Find the position of the pendulum at \( t = 4.25 \) seconds:
By substituting \( t = 4.25 \) into the distance function:
[tex]\[ d = 6 \cos(\pi \cdot 4.25) \][/tex]
The value of this is found to be approximately:
[tex]\[ d \approx 4.243 \text{ inches} \][/tex]
Thus, the completed solution is:
- The period is \( 2 \) seconds.
- The value of \( b \) is \( \pi \).
- The least positive value of \( t \) at which the pendulum is in the center is \( 0.5 \) seconds.
- The position of the pendulum at \( t = 4.25 \) seconds is \( 4.243 \) inches to the nearest thousandth.
The equation of the model remains:
[tex]\[ d = 6 \cos(\pi t) \][/tex]
The problem involves a pendulum with a horizontal distance traveled of 12 inches from right to left. This distance corresponds to the amplitude of the motion. Given that the pendulum takes 1 second to swing back from left to right, we understand that the complete period of the pendulum’s motion (one full cycle) is 2 seconds.
Given:
- The amplitude of the motion, which is the maximum distance from the center, is 6 inches. Thus, \( a = 6 \).
1. Determine the period:
The period is the time taken for one complete cycle of the pendulum. Since it takes 1 second to swing back from left to right, the total period (swinging both ways) is 2 seconds.
[tex]\[ \text{Period} = 2 \text{ seconds} \][/tex]
2. Determine the value of \( b \):
The value of \( b \) is derived from the period of the cosine function. The relationship is given by:
[tex]\[ b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{2} = \pi \][/tex]
3. Write the cosine function:
Combining the amplitude and the value of \( b \), the distance \( d \) of the pendulum from the center as a function of time \( t \) is given by:
[tex]\[ d = 6 \cos(\pi t) \][/tex]
4. Find the least positive value of \( t \) at which the pendulum is in the center:
The pendulum is in the center when \( d = 0 \). For the function \( d = 6 \cos(\pi t) \), the value of \( d \) is zero when:
[tex]\[ \pi t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \][/tex]
The least positive \( t \) corresponds to:
[tex]\[ t = \frac{1}{2} = 0.5 \text{ seconds} \][/tex]
5. Find the position of the pendulum at \( t = 4.25 \) seconds:
By substituting \( t = 4.25 \) into the distance function:
[tex]\[ d = 6 \cos(\pi \cdot 4.25) \][/tex]
The value of this is found to be approximately:
[tex]\[ d \approx 4.243 \text{ inches} \][/tex]
Thus, the completed solution is:
- The period is \( 2 \) seconds.
- The value of \( b \) is \( \pi \).
- The least positive value of \( t \) at which the pendulum is in the center is \( 0.5 \) seconds.
- The position of the pendulum at \( t = 4.25 \) seconds is \( 4.243 \) inches to the nearest thousandth.
The equation of the model remains:
[tex]\[ d = 6 \cos(\pi t) \][/tex]
