Mathematics, 21.10.2019 15:30 Tren17

1. the table shows the height, in centimeters, that a weight bouncing from a spring would achieve if there were no friction, for a given number of seconds.

2. how many seconds are required for the pendulum to swing from its position furthest to the right to its position furthest to the left?

3. how many degrees is the painted spot turning in 0.75 s?
enter your answer in the box.

4. what is the average daily maximum temperature in march?
round to the nearest tenth of a degree if needed.
use 3.14 for π .
enter your answer in the box

5. what is the greatest temperature the substance reached during the experiment?
round to the nearest tenth of a degree if needed.
use 3.14 for π .
enter your answer in the box.

1. the table shows the height, in centimeters, that a weight bouncing from a spring would achieve if

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Answers

Answer from: GiaTeyy6536

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;

$\begin{tabular}{|c|c|l|}\underline{Time (s)} &\underline{Height (cm)}&\underline{Position}\\&&\\1.5&0&Resting position\\3&25&Highest point in upward direction\\4.5&0&Resting position\\6.0&-25&Lowest point of weight (downward direction)\\7.5&0&Resting position\end{array}\right]$

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 = $\displaystyle \frac{T}{2} = \frac{2.5 \, s}{2} = \underline{1.25 \, s}$

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;

$\begin{tabular}{ccc}The angle rotated in 0.75 s = \end{array}\right] \dfrac{360^{\circ}}{1.5 \, s} \times 0.75 \, s = \underline{180^{\circ}}$

4. The function that gives the maximum temperature is presented as follows;

$\displaystyle f(x) = \mathbf{12.2 \cdot cos \left(\frac{\pi \cdot x}{6} \right) + 54.9}$

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;

$\displaystyle f(2) = 12.2 \cdot cos \left(\frac{\pi \times 2}{6} \right) + 54.9 = \mathbf{61}$

The maximum temperature in March f(2) = 61 °F

5. The function that gives the temperature of the substance is presented as follows;

$\displaystyle f(x) = \mathbf{-7.5 \cdot cos \left(\frac{\pi \cdot x}{30} \right) + 31.4}$

The value of x at the maximum temperature is given as follows;

$\displaystyle f'(x) = \frac{d}{dx} \left(-7.5 \cdot cos \left(\frac{\pi \cdot x}{30} \right) + 31.4\right) = \frac{\pi}{4} \cdot sin \left(\frac{\pi \cdot x}{30} \right) = 0$

Which gives;

$\displaystyle \frac{\pi \cdot x}{30} = 0$

∴ x = 0

$\displaystyle \frac{\pi \cdot x}{30} = \pi$

∴ x = 30

When x = 0, we get;

$\displaystyle f(0) = -7.5 \cdot cos \left(\frac{\pi \times 0}{30} \right) + 31.4 =23.9$

When x = 30, we get;

$\displaystyle f(30) = -7.5 \cdot cos \left(\frac{\pi \times 30}{30} \right) + 31.4 = \mathbf{38.9}$

The maximum temperature is therefore; 38.9 °C

Learn more about periodic function here:

link

Answer from: wmaingrette1

Problem 1

6

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At time 1.5s the height is 0 which is the resting position. The next time the height is 0 is when t = 4.5 which is a difference of 3 seconds (4.5 - 1.5 = 3). Double this to get 3*2 = 6

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Problem 2

1.25

-------------------------

Notice how the period of this graph is 2.5 seconds. It takes 2.5 seconds for the up and down pattern to repeat fully again. The easiest way to see this is from x = 0 to x = 2.5 the down and then up pattern does one full cycle, only to repeat again from x = 2.5 to x = 5, and so on.

Half of this cycle is the length of time it takes for the particle to go from the peak to the valley, or vice versa. Half of 2.5 is 1.25 (2.5/2 = 1.25). Why is it half? Because a full cycle takes place from one valley to the next neighboring valley. The same happens with the peaks as well.

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Problem 3

180 degrees

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A full cycle takes 1.5 seconds, which consists of the wheel rotating 360 degrees (a full revolution)
Half a cycle is 1.5/2 = 0.75 seconds which is half a revolution 360/2 = 180 degrees

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Problem 4

61

-------------------------

We plug in x = 2 since x = 0 corresponds to january, x = 1 corresponds to february, and so on

f(x) = 12.2*cos(pi*x/6) + 54.9
f(2) = 12.2*cos(pi*2/6) + 54.9
f(2) = 12.2*cos(3.14*2/6) + 54.9
f(2) = 12.2*cos(3.14/3) + 54.9
f(2) = 12.2*cos(1.04666666666667) + 54.9
f(2) = 12.2*0.5004596890082 + 54.9
f(2) = 6.10560820590003 + 54.9
f(2) = 61.0056082059
f(2) = 61

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Problem 5

38.9

-------------------------

The smallest cos(x) can be is -1. It doesn't matter if its cos(x) or cos(2x) or whatever the argument is. Therefore f(x) maxes out at f(x) = -7.5*(-1)+31.4 = 38.9

1. the table shows the height, in centimeters, that a weight bouncing from a spring would achieve if