Question

Given the trigonometric function
y = 120 sin ((Pi/6)*x - Pi/3) where x is in radians representing seconds
a._______________ is the amplitude of the
function.
b._______________radians is the period of the
function in seconds which is _________________hertz.(cycles per second)
c._______________radians is the phase shift(right or left

Answer 1

Given the trigonometric function

$y = 120 \sin \left( \frac{(P_i / 6)^* x - P_i / 3}{3} \right)$ where $x$ is in radians representing seconds

$$y = 120 \sin \left[ \frac{\pi}{6} x - \frac{\pi}{3} \right]$$

The general form of the sine function is:

$$y = A \sin (B x + C) + D$$

where:

$A$ is the amplitude of the function

The period of the function is: $T = \frac{2\pi}{B}$

The phase shift of the function is: $\frac{C}{B}$

a.

$A = 120$, therefore:

120 is the amplitude of the function.

b.

$B = \frac{\pi}{6}$, $T = \frac{2\pi}{B} = 12$ - period

numbers cycles per second: $n = \frac{1}{T} = \frac{1}{12}$ Hz

12 radians is the period of the function in seconds which is $\frac{1}{12}$ hertz.(cycles per second)

c.

$$\frac{C}{B} = -\frac{1}{2}$$

1/2 radians is the phase shift right, because phase shift with the sign “-”.