Sagot :
Answer:
Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called simple harmonic motion and it is important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.
The time it takes a pendulum to complete one whole back-and-forth swing is called the period of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: T=aL−−√. The proportionality constant, a, depends on the acceleration of gravity: a=2πg√. At sea level on Earth, acceleration of gravity is g=9.81 m/s2 (meters per second squared). Using this value of gravity, we find a=2.0 with units of sm−−√ (seconds divided by the square root of meters).
Up until the mid 20th century, all clocks used pendulums as their central time keeping component.
Real-World Application: Pendulums
Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?
The function for the period of a pendulum at sea level is T=2L−−√.
We start by making a table of values for this function:
L T=2L−−√
0 T=20–√=0
1 T=21–√=2
2 y=22–√=2.8
3 y=23–√=3.5
4 y=24–√=4
5 y=25–√=4.5
Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.
We can see from the graph that a length of approximately 14 meters gives a period of 1 second. We can confirm this answer by using our function for the period and plugging in T=1 second:
T=2L−−√⇒1=2L−−√
Square both sides of the equation:Solve for L:1=4LL=14 meters
Real-World Application: TV Screens
“Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is 43 the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of 180 in2?
Let d= length of the diagonal, x= width
Then 4 × height = 3 × width
Or, height = 34x.
The area of the screen is: A= length × width or A=34x2
Find how the diagonal length relates to the width by using the Pythagorean theorem:
x2+(34x)2x2+916x22516x2=d2=d2=d2⇒x2=1625d2⇒x=45d
Therefore, the diagonal length relates to the area as follows: A=34(45d)2=34⋅1625d2=1225d2.
We can also flip that around to find the diagonal length as a function of the area: d2=2512A or d=523–√A−−√.
Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values:
A d=523–√A−−√
0 0
25 7.2
50 10.2
75 12.5
100 14.4
125 16.1
150 17.6
175 19
200 20.4
From the graph we can estimate that when the area of a TV screen is 180 in2 the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in A=180 into the formula that relates the diagonal to the area: d=523–√A−−√=523–√180−−−√=19.4 inches.
Radicals often arise in problems involving areas and volumes of geometrical figures.
Real-World Application: Pool Dimensions
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.
Make a sketch:
Let x= the width of the pool. Then:
Area = length × width
Combined length of pool and walkway =2x+2
Combined width of pool and walkway =x+2
Area=(2x+2)(x+2)
Since the combined area of pool and walkway is 400 ft2 we can write the equation
(2x+2)(x+2)=400
Multiply in order to eliminate the parentheses:Collect like terms:Move all terms to one side of the equation:Divide all terms by 2:Use the quadratic formula:2x2+4x+2x+42x2+6x+42x2+6x−396x2+3x−198xxxx=400=400=0=0=−b±b2−4ac−−−−−−−√2a=−3±32−4(1)(−198)−−−−−−−−−−−−−√2(1)=−3±801−−−√2=−3±28.32=12.65 feet
(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)
So the dimensions of the pool are: length=12.65 and width=25.3 (since the width is 2 times the length)
That means that the area of just the pool is A=12.65⋅25.3→320ft2
Check by plugging the result in the area formula:
Area =(2(12.65)+2)(12.65+2)=27.3⋅14.65=400 ft2.