Answer :
To find the trigonometric function that models the distance [tex]\( H \)[/tex] from the pendulum's bob to the wall after [tex]\( t \)[/tex] seconds, given the information:
- Period [tex]\( T = 0.8 \)[/tex] seconds
- Amplitude [tex]\( A = 6 \)[/tex] cm
- Midline [tex]\( H = 15 \)[/tex] cm
- At [tex]\( t = 0.5 \)[/tex] seconds, the bob is at its midline and moving towards the wall.
We can proceed as follows:
### Step 1: Determine the Angular Frequency
The angular frequency [tex]\( \omega \)[/tex] is calculated using the formula:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
Given the period [tex]\( T = 0.8 \)[/tex] seconds:
[tex]\[ \omega = \frac{2\pi}{0.8} = 7.853981633974483 \][/tex]
### Step 2: Formulate the Function Based on Trigonometric Characteristics
The displacement of the pendulum from the midline can be represented by a cosine function because at [tex]\( t = 0.5 \)[/tex] seconds, the bob is at the midline and moving towards the wall, which aligns well with the behavior of the cosine function offset by [tex]\(\pi/2\)[/tex] phase shift:
[tex]\[ H(t) = A \cos(\omega t + \phi) + \text{midline} \][/tex]
### Step 3: Determine the Phase Shift [tex]\(\phi\)[/tex]
Since the bob is at the midline at [tex]\( t = 0.5 \)[/tex] seconds and moving towards the wall, let's set:
[tex]\[ \cos(\omega \cdot t + \phi) = 0 \][/tex]
At [tex]\( t = 0.5 \)[/tex] seconds:
[tex]\[ \cos(7.853981633974483 \cdot 0.5 + \phi) = 0 \][/tex]
Solving for [tex]\(\phi\)[/tex]:
[tex]\[ 7.853981633974483 \cdot 0.5 + \phi = \frac{\pi}{2} \][/tex]
[tex]\[ \phi = \frac{\pi}{2} - 3.9269908169872415 = -1.5707963267948966 \][/tex]
### Step 4: Write the Final Equation
Now we combine all the values into the cosine function to represent the distance [tex]\( H(t) \)[/tex].
Thus, the distance [tex]\( H(t) \)[/tex] from the pendulum's bob to the wall is:
[tex]\[ H(t) = 6 \cos(7.853981633974483 \cdot t - 1.5707963267948966) + 15 \][/tex]
Therefore, the formula for [tex]\( H(t) \)[/tex] is:
[tex]\[ H(t) = 6 \cos(7.853981633974483 t - 1.5707963267948966) + 15 \][/tex]
- Period [tex]\( T = 0.8 \)[/tex] seconds
- Amplitude [tex]\( A = 6 \)[/tex] cm
- Midline [tex]\( H = 15 \)[/tex] cm
- At [tex]\( t = 0.5 \)[/tex] seconds, the bob is at its midline and moving towards the wall.
We can proceed as follows:
### Step 1: Determine the Angular Frequency
The angular frequency [tex]\( \omega \)[/tex] is calculated using the formula:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
Given the period [tex]\( T = 0.8 \)[/tex] seconds:
[tex]\[ \omega = \frac{2\pi}{0.8} = 7.853981633974483 \][/tex]
### Step 2: Formulate the Function Based on Trigonometric Characteristics
The displacement of the pendulum from the midline can be represented by a cosine function because at [tex]\( t = 0.5 \)[/tex] seconds, the bob is at the midline and moving towards the wall, which aligns well with the behavior of the cosine function offset by [tex]\(\pi/2\)[/tex] phase shift:
[tex]\[ H(t) = A \cos(\omega t + \phi) + \text{midline} \][/tex]
### Step 3: Determine the Phase Shift [tex]\(\phi\)[/tex]
Since the bob is at the midline at [tex]\( t = 0.5 \)[/tex] seconds and moving towards the wall, let's set:
[tex]\[ \cos(\omega \cdot t + \phi) = 0 \][/tex]
At [tex]\( t = 0.5 \)[/tex] seconds:
[tex]\[ \cos(7.853981633974483 \cdot 0.5 + \phi) = 0 \][/tex]
Solving for [tex]\(\phi\)[/tex]:
[tex]\[ 7.853981633974483 \cdot 0.5 + \phi = \frac{\pi}{2} \][/tex]
[tex]\[ \phi = \frac{\pi}{2} - 3.9269908169872415 = -1.5707963267948966 \][/tex]
### Step 4: Write the Final Equation
Now we combine all the values into the cosine function to represent the distance [tex]\( H(t) \)[/tex].
Thus, the distance [tex]\( H(t) \)[/tex] from the pendulum's bob to the wall is:
[tex]\[ H(t) = 6 \cos(7.853981633974483 \cdot t - 1.5707963267948966) + 15 \][/tex]
Therefore, the formula for [tex]\( H(t) \)[/tex] is:
[tex]\[ H(t) = 6 \cos(7.853981633974483 t - 1.5707963267948966) + 15 \][/tex]