Question

The current \(I\), in amperes, flowing through an ac (alternating current) circuit at time \(t,\) in seconds, is $$ I(t)=120 \sin \left(30 \pi t-\frac{\pi}{3}\right) \quad t \geq 0 $$ What is the period? What is the amplitude? What is the phase shift? Graph this function over two periods.

Short answer

Amplitude: 120, Period: \( \frac{1}{15} \) seconds, Phase Shift: \( \frac{1}{90} \) seconds.

Step-by-step solution

Identify the General Form

The given function is \[ I(t)=120 \, \sin \left(30 \pi t - \frac{\pi}{3} \right) \] . This can be compared to the general form of a sinusoid, which is \[ I(t) = A \sin (Bt + C) \] .

Determine the Amplitude

The amplitude of a sinusoidal function is the coefficient in front of the sine function. Here, the amplitude \(A\) is 120.

Determine the Period

The period of a sinusoidal function is given by: \[ T = \frac{2\pi}{|B|} \]. In this function, \( B = 30 \pi \). Therefore, the period \( T \) is: \[ T = \frac{2\pi}{30\pi} = \frac{1}{15} \] seconds.

Determine the Phase Shift

The phase shift can be found by setting the argument of the sine function to zero: \[ 30 \pi t - \frac{\pi}{3} = 0 \Rightarrow t = \frac{1}{90} \]. So, the phase shift is \( \frac{1}{90} \) seconds.

Graph the Function over Two Periods

To graph the function over two periods, we need to consider \(2\times T\), which is \(2 \times \frac{1}{15} = \frac{2}{15} \) seconds. Plot the function from \( t = 0 \) to \( t = \frac{2}{15} \). The graph will show two complete oscillations of the sine wave.

Key concepts

amplitude of sine function

The amplitude of a sine function is a measure of its peak value, or how far the wave oscillates above and below its central axis.
For the given function, $$ I(t)=120 \, \sin \left(30 \pi t - \frac{\pi}{3} \right) $$, the amplitude is the coefficient in front of the sine term, which is 120. This tells us that the maximum current reaching through the circuit is 120 amperes, and the minimum is -120 amperes.
To identify the amplitude in general form, compare it to $$ I(t) = A \sin (Bt + C)$$ where A represents the amplitude. In this exercise, 120 is the amplitude.

period of sine function

The period of a sine function is the length of time it takes for the wave to complete one full cycle.
For our given function \( I(t)=120 \sin \left(30 \pi t - \frac{\pi}{3} \right) \), we determine the period using the general form \( I(t) = A \sin (Bt + C) \) where the period T is given by: $$ T = \frac{2\pi}{|B|}$$.
In this case, B is \(30 \pi\). Thus, the period is: $$ T = \frac{2\pi}{30\pi} = \frac{1}{15} $$ seconds.
This means the function repeats itself every \(\frac{1}{15}\) seconds.

phase shift of sine function

Phase shift refers to the horizontal shift of the sine function along the time axis.
To find the phase shift of our function $$ I(t)=120 \sin \left(30 \pi t - \frac{\pi}{3} \right) $$, observe the argument of the sine function. The phase shift is derived from the term adding or subtracting from \( Bt \) inside the sine function.
Setting the argument of the sine function to zero gives: $$ 30 \pi t - \frac{\pi}{3} = 0 \Rightarrow t = \frac{1}{90} $$.
Thus, our phase shift is \( \frac{1}{90} \) seconds to the right.

graphing sinusoidal functions

Graphing sinusoidal functions involves plotting the wave over one or more periods to see the repeated oscillations.
For the given function $$ I(t)=120 \sin \left(30 \pi t - \frac{\pi}{3} \right) $$, we graph it over two periods. Since one period is \( \frac{1}{15} \) seconds, two periods will be \( \frac{2}{15} \) seconds.
Start by plotting key points such as the maximum, minimum, and points where the wave crosses the time axis. These points often occur at intervals of \( \frac{T}{4} \) where T is the period.
The steps to create the graph:

From \( t = 0 \) to \( t = \frac{2}{15}\), identify points at \(\frac{ \frac{1}{15}}{4}\) intervals.

Plot where the wave reaches its peaks (120A), troughs (-120A), and crosses the axis (0A).

Connect these points smoothly to reflect the sinusoidal nature, showing two full cycles.