The motion of the particles are periodic motion that that oscillates about a
given point position.
- 1. The time it takes is 6.0 seconds.
- 2. The time it takes to swing from right to left is 1.25 seconds.
- 3. At 0.75 s, the painted spot has turned 0.75 seconds.
- 4. The maximum temperature in March is 61 °F.
- 5. The greatest temperature the substance reached is 38.9 °C.
Reasons:
1. The information in the table is presented as follows;
[tex]\begin{tabular}{|c|c|l|}\underline{Time (s)} &\underline{Height (cm)}&\underline{Position}\\&&\\1.5&0&Resting position\\3&25&Highest point in upward direction\\4.5&0&Resting position\\6.0&-25&Lowest point of weight (downward direction)\\7.5&0&Resting position\end{array}\right][/tex]
From the above table, we have the time it takes the weight to start from the
resting position, bounce and return from the upward direction, passing
through the resting position to the downward direction and back to the
resting position, (one period), T = 7.5 s - 1.5 s = 6.0 s
2. The time it takes for the pendulum to swing from the furthest position on
the to the right back to the furthest position on the right = 1 period, T
From the graph, one period, the time to complete one cycle, T = 2.5 s
Therefore, the time it takes the pendulum to move from the furthest position on the left to the furthest position on the right = [tex]\displaystyle \frac{T}{2} = \frac{2.5 \, s}{2} = \underline{1.25 \, s}[/tex]
3. From the graph, one period, which is the time it takes to complete one cycle, T = 1.5 s
One cycle = A rotation of 360°
Therefore;
It takes 1.5 s to rotate 360°, which gives;
[tex]\begin{tabular}{ccc}The angle rotated in 0.75 s = \end{array}\right] \dfrac{360^{\circ}}{1.5 \, s} \times 0.75 \, s = \underline{180^{\circ}}[/tex]
4. The function that gives the maximum temperature is presented as follows;
[tex]\displaystyle f(x) = \mathbf{12.2 \cdot cos \left(\frac{\pi \cdot x}{6} \right) + 54.9}[/tex]
Where;
f(x) = The temperature in °F
x = The month of the year
At x = 0, the month is January
Therefore;
January = 0
February = 1
March = 2
The temperature in March is therefore;
[tex]\displaystyle f(2) = 12.2 \cdot cos \left(\frac{\pi \times 2}{6} \right) + 54.9 = \mathbf{61}[/tex]
The maximum temperature in March f(2) = 61 °F
5. The function that gives the temperature of the substance is presented as follows;
[tex]\displaystyle f(x) = \mathbf{-7.5 \cdot cos \left(\frac{\pi \cdot x}{30} \right) + 31.4}[/tex]
The value of x at the maximum temperature is given as follows;
[tex]\displaystyle f'(x) = \frac{d}{dx} \left(-7.5 \cdot cos \left(\frac{\pi \cdot x}{30} \right) + 31.4\right) = \frac{\pi}{4} \cdot sin \left(\frac{\pi \cdot x}{30} \right) = 0[/tex]
Which gives;
[tex]\displaystyle \frac{\pi \cdot x}{30} = 0[/tex]
∴ x = 0
[tex]\displaystyle \frac{\pi \cdot x}{30} = \pi[/tex]
∴ x = 30
When x = 0, we get;
[tex]\displaystyle f(0) = -7.5 \cdot cos \left(\frac{\pi \times 0}{30} \right) + 31.4 =23.9[/tex]
When x = 30, we get;
[tex]\displaystyle f(30) = -7.5 \cdot cos \left(\frac{\pi \times 30}{30} \right) + 31.4 = \mathbf{38.9}[/tex]
- The maximum temperature is therefore; 38.9 °C
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