question 14 (4 points)\nappl: a water wheel has a radius of 2 m and 0.5 m of the wheel is submerged below…

question 14 (4 points)\nappl: a water wheel has a radius of 2 m and 0.5 m of the wheel is submerged below the water. it completes a complete rotation every 30 seconds.\na) sketch the graph of the wheel over two cycles starting at the lowest point.\nb) determine a sine function that models the height, h, in metres, relative to the water as a function of the time, t, in seconds.\n4

question 14 (4 points)\nappl: a water wheel has a radius of 2 m and 0.5 m of the wheel is submerged below the water. it completes a complete rotation every 30 seconds.\na) sketch the graph of the wheel over two cycles starting at the lowest point.\nb) determine a sine function that models the height, h, in metres, relative to the water as a function of the time, t, in seconds.\n4

Answer

Explanation:

Step1: Analyze the amplitude

The radius of the water - wheel is $r = 2$m and $0.5$m is submerged. The amplitude $A$ of the sine - function is the maximum displacement from the equilibrium position. The equilibrium position is $0.5$m above the water level. The amplitude $A=2$.

Step2: Determine the period

The wheel completes a full rotation in $T = 30$ seconds. The formula for the angular frequency $\omega$ is $\omega=\frac{2\pi}{T}$. Substituting $T = 30$ seconds, we get $\omega=\frac{2\pi}{30}=\frac{\pi}{15}$.

Step3: Find the vertical shift

The center of the wheel is $2 - 0.5=1.5$m above the water level. So the vertical shift $D = 1.5$.

Step4: Consider the phase shift

Since we start at the lowest point, the phase shift $C$ is such that the sine - function is at its minimum at $t = 0$. The general form of a sine function is $h(t)=A\sin(\omega(t - C))+D$. For a sine function $y = A\sin(\omega t+\varphi)$ (equivalent to $y = A\sin(\omega(t-\frac{-\varphi}{\omega}))$), when starting at the minimum, $\varphi=-\frac{\pi}{2}$. Since $\omega=\frac{\pi}{15}$, the phase - shift $C=\frac{15}{2}=7.5$ seconds. The sine function is $h(t)=2\sin(\frac{\pi}{15}(t - 7.5))+1.5$.

For part (a):

  1. Mark the water level as the $x$ - axis.
  2. The lowest point of the wheel is at $y=- 0.5$ (since $0.5$m is submerged) and the highest point is at $y = 3.5$ (radius of $2$m and center at $1.5$m above water).
  3. The period is $T = 30$ seconds. Start at the lowest point $(0,-0.5)$. After $7.5$ seconds, the wheel is at the water level (equilibrium), after $15$ seconds, it is at the highest point $(15,3.5)$, after $22.5$ seconds, it is back at the water level, and after $30$ seconds, it is back at the lowest point. Repeat for the second cycle from $t = 30$ to $t = 60$ seconds.

Answer:

a) Sketch as described above. b) $h(t)=2\sin(\frac{\pi}{15}(t - 7.5))+1.5$