We are told that an audiometer produced a pure tone with a frequency of

1000

hertz. We want to write a sine model that gives us the pressure

P

as a function of the time

t .

To begin, we will recall the general form of a sine function.

P = a sin b t

In this equation,

a

and

b

are non-zero real numbers. We know that an amplitude of basic sine function is

1 .

In this case we are given that the maximum pressure

P

is

2

millipascals. This means that the amplitude of our function is

2 .

In the general form of a sine function,

a

represents change of amplitude of the graph. Therefore, in our case we can set

a = 2 .

P = 2 sin b t

Now, we need to find

b .

We know that the frequency and the period are inversely proportional. In our case, the frequency is equal to

1000

hertz and the period is

\frac{2 π}{b} .

We can substitute these information into the relationship involving frequency and period.

frequency = \frac{1}{period} \Rightarrow 1000 = \frac{1}{\frac{2 π}{b}}

Let's solve this equation for

b .

1000 = \frac{1}{\frac{2 π}{b}}

DivOneByFracD

$\frac{1}{a / b} = \frac{b}{a}$

1000 = \frac{b}{2 π}

▼

Solve for

b

MultEqn

$LHS \cdot 2 π = RHS \cdot 2 π$

2 π (1000) = b

RearrangeEqn

Rearrange equation

b = 2 π (1000)

Multiply

b = 2000 π

Now that we have

a

and

b,

we can complete the equation for the maximum pressure.

P = 2 sin 2000 π t

To graph this model, we need to identify the points where intercepts, maximum values, and minimum values occur. To do so, we will substitute

a = 2

and

b = 2000 π

into the known formulas for the key points of a sine function. Let's show this in the table!

	Formula	Substitution	Decimal Form
Intercepts	$(0, 0),$ $(\frac{1}{2} \cdot \frac{2 π}{b}, 0),$ $(\frac{2 π}{b}, 0)$	$(0, 0),$ $(\frac{1}{2} \cdot \frac{2 π}{2 0 0 0 π}, 0),$ $(\frac{2 π}{2 0 0 0 π}, 0)$	$(0, 0),$ $(0.0005, 0),$ $(0.001, 0)$
Maximum	$(\frac{1}{4} \cdot \frac{2 π}{b}, a)$	$(\frac{1}{4} \cdot \frac{2 π}{2 0 0 0 π}, 2)$	$(0.00025, 2)$
Minimum	$(\frac{3}{4} \cdot \frac{2 π}{b}, - a)$	$(\frac{3}{4} \cdot \frac{2 π}{2 0 0 0 π}, - 2)$	$(0.00075, - 2)$

We can graph the function by plotting calculated points and connecting them with a smooth curve. Let's do it!

We draw a graph of a function of maximum pressure $P$ produced by pure tone with a frequency of $1000$ hertz. Finally, we can compare it with the original function $P = 2 sin 2000 π t$ to see how the graph of a function changed, when we changed a frequency.

We changed the frequency from

2000

hertz into

1000

hertz, which means that it became two times smaller. As a result, we can see that the period of a graph became two times greater. In other words the graph is horizontally stretched two times. This does not surprise us, let's again look at the relationship involving frequency and period.

frequency = \frac{1}{period}

Their inverse proportionality implies that when the frequency gets two times smaller, then the period gets two times greater. This is the same conclusion as we observed at the graph.

Exercise