TRIG WHEEL 2
Let's lift the wheel in Trig 1 vertically out of the water so that it's 1 m clear. Now increase the speed of the wheel so that it completes 2 revolutions in 60 seconds. So the wheel is 8 m in diameter and its lowest point is 1 m above the 'x' axis. The maximum point will be at 9 m.
Write the equation again starting from the same point as before, a sine wave function with amplitude (A), frequency (B), shift (C) (again this will be zero) and vertical shift (D).
Sketch a diagram to get a feel for the problem like above.
Write the equation again starting from the same point as before, a sine wave function with amplitude (A), frequency (B), shift (C) (again this will be zero) and vertical shift (D).
Sketch a diagram to get a feel for the problem like above.
Amplitude (A)
The amplitude is the same as in the previous example which was the radius of the circle or 4 m.
Frequency (B)
The frequency B is equal to the distance divided by the time for one period, which is:
Horizontal (phase) shift (C)
As before we are dealing with a normal sine wave so there is no shift to be concerned with.
Vertical Shift (D)
Because the wheel has been lifted or 'shifted' vertically, the axle (circle origin) is now at a height of 5 m. So D = 5 and the equation for the function becomes.