The graph hits [tex]\fbox{\begin\\\ \dfrac{10}{T}+2\\\end{minispace}}[/tex] times over 2 feet for [tex]a>2[/tex].
Further explanation:
The height of the wave is given by the equation as follows:
[tex]y=asin\left(\dfrac{2\pi t}{T}\right)[/tex] ......(1)
Here, [tex]a[/tex] is amplitude, [tex]T[/tex] is period of wave in second and [tex]t[/tex] time in seconds.
The height [tex]y[/tex] of the wave is given as 2 feet and time [tex]t[/tex] is given as 5 seconds.
Substitute 2 for [tex]y[/tex] and 5 for [tex]t[/tex] in equation (1).
[tex]2=asin\left(\dfrac{2\pi \times5}{T}\right)\\2=asin\left(\dfrac{10\pi}{T}\right)\\\dfrac{2}{a}=sin\left(\dfrac{10\pi}{T}\right)[/tex]
The above eqution is valid only for [tex]a\geq 2[/tex] because the maximum value of the term [tex]sin(10\pi /T)[/tex] is 1.
If [tex]T[/tex] is the time period then in [tex]T[/tex] seconds the graph will hit at least 2 times over 2 feet for [tex]a>2[/tex].
T seconds[tex]\rightarrow[/tex]2 hits
1 seconds [tex]\rightarrow[/tex] [tex]\dfrac{2}{T}[/tex] hits
5 seconds [tex]\rightarrow\dfrac{2\times5}{T}[/tex]
5 seconds [tex]\rightarrow[/tex] [tex]\dfrac{10}{T}[/tex]
If [tex]T[/tex] is time period in 5 seconds then the graph will hit [tex][10/T][/tex] times in interval 0 to [tex]2\pi[/tex].
Thus, the graph hits [tex]\fbox{\begin\\\ \dfrac{10}{T}+2\\\end{minispace}}[/tex] times over 2 feet for [tex]a>2[/tex].
Learn more:
1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)
https://brainly.com/question/1332667
2. Which is the graph of f(x) = (x – 1)(x + 4)?
https://brainly.com/question/2334270
Answer details:
Grade: High school.
Subjects: Mathematics.
Chapter: function.
Keywords: Function, wave equation, height, amplitude, equation, period, periodic function, y=asin(2pit/T), frequency, magnitude, feet, height, time period, seconds, inequality, maximum value, range, harmonic motion, oscillation, springs, strings, sonometer.
