Title: Default for 5.3 Oscillating Functions and Modelling
Course: HELM Workbooks 2023-2024

Exercise

Write down the amplitude, period and frequency of the function \[h = [value]\sin([value]t + [value])\]

\(\text{amplitude} \quad A = \)

\(\text{period}, \qquad T =\) (Keep 2 decimal places)

\(\text{frequency} \quad f = \) (Keep 2 decimal places)
Feedback: We are given the function \[h = [value]\sin([value]t + [value])\] The amplitude is \(A = [value]\) The period is \(T = \dfrac{2\pi}{[value]} = [value] =[value]\) The frequency is \(f = \dfrac{1}{T} = [value] = [value]\)

Q: Exercise

The figures below show plots of the functions
\((\text{a})\,\,y=\sin x,\quad (\text{b})\,\,y=2\sin x,\quad (\text{c})\,\,y=\cos x,\quad\text{and}\quad (\text{d})\,\,y=\cos\dfrac{x}{2}\)

[Image]
[Image]

State the correct amplitude, frequency, and period for the four functions:

1. The amplitude is [...], the frequency is [...]\(\pi\), and the period if [...]\(\pi\).
2. The amplitude is [...], the frequency is [...]\(\pi\), and the period if [...]\(\pi\).
3. The amplitude is [...], the frequency is [...]\(\pi\), and the period if [...]\(\pi\).
4. The amplitude is [...], the frequency is [...]\(\pi\), and the period if [...]\(\pi\).
Also offered: 3; 4; 3/4; 2; 6; 3
A: 1; 1/2; 2; 1; 1/2; 2; 2; 1/2; 2; 1; 1/4; 4
Feedback: In general, functions of the form \(y=a\sin(bx),\quad\text{and}\quad y=a\cos(bx)\) Have amplitude equal to \(a\), frequency equal to \(\dfrac{b}{2\pi}\), and period equal to \(\dfrac{2\pi}{b}\).

The peak levels of the graph correspond to times when the sine function has the value 1. The lowest points correspond to times when the sine function is \(-1\). At these times the arguments of the sine function (i.e. \(2.7t+8.5\)) are an odd number of \(\pi/2\) starting at \(3\pi /2\) for the first low tide.

So far all of this may be deduced from the general form \(y = a\)sin\((bx)\) and from the modelling context. However there is an additional term in the function being considered here. This is a constant \(8.5\) within the sine function. When \(t = 0\) the presence of this constant means that the intercept on the height axis is 3.2sin\((8.5)\) = 2.56, implying that the water level is 2.56 m above the mean value at the start of timing. The constant 8.5 has displaced the sine curve sideways. This constant is known as the phaseof the function. Phase is measured in radians as it is an angle.

As remarked earlier, at \(t = 0\), this function has the value \(3.2\sin (8.5)\). Since \(\sin (8.5) = \sin(8.5 - 2\pi) \approx \sin (2.2168)\), we can replace the constant 8.5 by 2.2168 without altering the values on the graph. This means that the function

\(\qquad h = 3.2\sin (2.74t + 2.2168)\)

does just as well as the original function in representing the tidal variation in the harbour. We now rewrite this latest form of the function, representing the variation of water level in the harbour, so that time is measured in hours rather than in six-hourly invervals. The effect of changing the units of time to hours from 6 hours is to decrease the coefficient of \(t\) in the sine function by a factor of 6, so that the new function is

\(\qquad h = 3.2 \sin (0.45t + 2.2168).\)

See the figure below

[Image]

We can use the latest form of the function to calculate the time of the first low tide assuming that \(t = 0\) corresponds to midnight.

At the first low tide, \(h = -3.2\)\(\quad\)and\(\quad \sin(0.45t + 2.2168) = -1,\)

Using the fact that \(\sin(\dfrac{3\pi}{2})=-1\), we have

\(\qquad 0.45t + 2.2168 = 3\pi /2\),\(\quad\)giving\(\quad\)\(t = 5.5458=5.55\)\(\quad\)to \(2\) d.p.

so the first low tide is at 5:30 a.m.

There are various rules connected with sine and cosine functions that can be summarised at this point.

(1)\(\quad\)Placing a multiplier before \(\sin x\) or \(\cos x\) (e.g. \(2\sin x\)) changes the amplitude without changing the period.

(2)\(\quad\)Placing a multiplier before \(x\) in \(\sin x\) or \(\cos x\), (e.g. \(\sin 3x\)), changes the period or frequency without changing the amplitude.

(3)\(\quad\)As with any function, the addition of a constant (e.g. \(4+\sin x\)) raises or lowers the whole graph of the sine or cosine function. It alters the mean value without changing the amplitude.

(4)\(\quad\)Changing the sign within a cosine function has no effect, (e.g. \(\cos (-x)\) = \(\cos x\)).

(5)\(\quad\)Changing the sign within a sine function changes the sign of the function, (e.g. \(\sin (-x)\) = \(-\sin x\)).

(6)\(\quad\)Placing a constant or altering the constant \(b\) in \(\sin (ax + b)\) or \(\cos (ax + b)\) changes the phase and shifts the sine or cosine function along the \(x\)-axis.

Oscillating functions: amplitude, period and frequency

Particular types of periodic functions that are especially important in engineering are the sine and cosine functions. These are possible choices when modelling behaviour that involves oscillation or motion in a circle. The usefulness of these functions is rather limited if we confine our attention only to \(\sin(x)\) and \(\cos(x)\). Use of functions such as \(3\sin(2x)\), \(5\cos(3x)\) and so on, and other functions made up of sums of functions of this type, enables the modelling of a great variety of situations where the quantity being modelled is known to change in a periodic way. Here we will examine the behaviour of sine and cosine functions and consider a modelling context where choice of a sine function is appropriate. Figure 6 shows how the terms amplitude, period and frequency are defined with respect to a general sinusoid (the name for any general sine or cosine function).

[Image]

Defining amplitude and period for a sinusoid

The amplitude represents the difference between the maximum (or minimum) value of a sinusoidal function and its mean value (which is zero in Figure 6). The frequency represents the number of complete cycles of the function in each unit change in \(x\). The period is such that \(f(x + T) = f(x)\) for all \(x\), e.g. for \(\sin x\), \(T=2\pi\).

Note that the cosine functions \(\cos nx\) have the same shape as the sine functions \(\sin nx\) but, at \(x = 0\), the cosine functions have a peak or maximum, whereas the sine functions have the value zero, which is the mean value for both of these functions. Indeed the graph of \(y = \cos x\) is exactly like that for \(y= \sin x\) with all the \(x\) values displaced by \(\pi /2\). More general forms of sine and cosine function are given by \(y\) = \(a\sin (bx)\), and \(y\) = \(a\cos(bx)\) where \(a\) and \(b\) are arbitary constants. These are functions with frequency \(\dfrac{b}{2\pi}\), period \(\dfrac{2\pi}{ b}\) and amplitude \(a\). The peak values of the sine functions occur at \(x\) values equal to \(\dfrac{\pi}{2}\), \(\dfrac{5\pi }{2}\), \(\dfrac{9\pi }{2}\) etc. The minimum values occur at \(x\) values equal to \(\dfrac{3\pi }{2}\), \(\dfrac{7\pi }{2}\), \(\dfrac{11\pi }{2}\) etc. When the period is measured in seconds, frequency is measured in cycles per second or Hz which has units of 1/time.

Oscillating functions: modelling tides

We consider how the function

\[\qquad h = 3.2 \sin (2.7t + 8.5)\]

might be used to model the rise and fall of the tide in a harbour. The figure below shows a graph of this function for \((0\ge t \ge 5)\).

[Image]

We consider some aspects of this graph and model. It seems reasonable to suppose that the tide creates an oscillation of the water level in the harbour of \(h\)m about some mean value represented on the graph by \(h = 0\). There seems to be a low tide near \(t = 1\) and another low tide just after \(t = 3\). Since we expect intervals of 12 to 14 hours between low tides around the U.K., this suggests that time in this graph is specified in 6-hour intervals.

Exercise

In this question we will consider how the function

\[h = 3.2\sin(0.45t + 2.2168)\]

might be used to model the rise and fall of the tide in a harbour. Assume that \(t=0\) corresponds to midnight. Calculate the time \(t\), of the first high tide after midnight

\(t=\)

Give your answer accurate to 2 d.p. The figure below may help you to visualise this question

[Image]
Feedback: The function representing the tidal variation in the harbour is \[h = 3.2\sin(0.45t + 2.2168)\] At the first high tide, \(h=3.2\) and \(\sin(0.45t+2.2168)=1\), so \(0.45t+2.2168=5\pi/2\) giving \(t=12.5271\approx12.53\) so the first high tide is at half past midday.

Exercise

Write down a function relating water level \(L\) (meters) in a harbour, to time \(t\) (hours), starting when the level is equal to the mean level of \([value] \, \textrm{m}\), that has an amplitude of \([value] \, \textrm{m}\) and has a period of twelvehours.

\(L= \)
Feedback: In the general form \(y=a\sin(bt+c)+d\), the phase is \(c=0\), the period is equal to \(12\), therefore \(b=\dfrac{2\pi}{12}=\dfrac{\pi}{6}\), the amplitude is \(a=[value]\), and the mean value is \(d=[value]\). Hence \(L=\displaystyle [value] \qquad (t \geq 0).\)

Exercise

The figure below shows a typical variation of the depth \(d\) (metres) of water in a particular harbour with time \(t\) (hours) as the depth changes with the tide.

[Image]

You are told that the equation for this function has the form

\(d=a+b\cos\left(\dfrac{c\pi t}{25}\right)\)

Determine \(a\), \(b\), and \(c\).

\(a=\)

\(b=\)

\(c=\)
Feedback: SOLUTION: By inspection \(a=5\), and \(b=3\). The period \(T=12.5=\dfrac{2\pi}{(c\pi/25)}\) so \(c=4\).

Exercise

Write down the amplitude, period, frequency and phase of the following functions.

Write period and frequency as multiples of \(\pi.\)

1. \(y = [value]\)
\(\text{amplitude} \quad A = \)
\(\text{period}, \qquad T =\)
\(\text{frequency} \quad f = \)
2. \(y = [value]\)
\(\text{amplitude} \quad A = \)
\(\text{period}, \qquad T =\)
\(\text{frequency} \quad f = \)
\(\text{phase} \quad \phi = \)
Feedback: 1. For the function \(y = [value]\) The amplitude is \(A = [value]\) The period is \(T = \dfrac{2\pi}{[value]} = \displaystyle [value]\) The frequency is \(f = \dfrac{1}{T} = \displaystyle [value]\) 2. For the function \(y = [value]\) The amplitude is \(A = [value]\) The period is \(T = \dfrac{2\pi}{[value]} = \displaystyle[value]\) The frequency is \(f = \dfrac{1}{T} = \displaystyle [value]\) The phase is \( \phi = [value]\)

Example

The diagram shows a typical variation of the depth (\(d\) metres) of water in a particular harbour with time (\(t\) hours) as the depth changes with the tide.

[Image]

1. Find a suitable equation for the curve in the diagram
2. A boat enters the harbour in late morning on a day when the high tide is at 2 p.m. The boat needs a water depth of \(4\,\textrm{m}\) to sail safely. What advice would you give to its pilot about when to leave the harbour if the boat is not to be forced to wait in the harbour through the evening low tide?
3. State two modelling assumptions you have made

Solution

1. The equation is of the form
\( \quad h = a+b\cos(\omega t) \quad \) (or \(h= a+b\sin \left(\omega t + \frac{\pi}{2}\right)\) )
By inspection, \( a= 5 \) and \(b=3.\)
The period \( T = 12.5 = \dfrac{2 \pi}{\omega}\) so \(\omega = \dfrac{4 \pi}{25} = 0.502655\)
so the equation of the curve is \(h = 5+3 \cos \left(\dfrac{4 \pi}{25}t \right).\)
2. Put \( h= 4 \) into the equation:
\( \quad 4 = 5+3\cos \left(\dfrac{4 \pi}{25}t \right) \) implying \( -\dfrac{1}{3}= \cos\left(\dfrac{4 \pi}{25}t \right) \)
Now, inverting the cosine:
\( \dfrac{4 \pi}{25}t = \cos^{-1} \left(-\dfrac{1}{3}\right) = 1.91063\) giving \(t = 3.80108\) hours.
So the advice to the pilot should be that he needs to be clear of the harbour by 5:45 pm at the very latest - and that he should allow a safety margin.
3. Assumptions likely are:
The tide on the day in question is typical.
No waves.
A sinusoidal function accurately models the effect of the tide on sea level.