Periods & Amplitudes (Cambridge International IGCSE Maths)

Periods & Amplitudes

What are oscillating graphs?

• The sine and cosine graphs are oscillating graphs

  • Both graphs oscillate between a maximum and a minimum point
  • A cycle is completed when the graph returns an initial starting point
  • The cycle is repeated indefinitely in both the positive and negative x directions

What are the features of oscillating graphs?

• An oscillating graph has the following key features:

  • The amplitude is half the vertical distance from the maximum point to the minimum point

    • This is the distance from the equationhorizontal line of symmetry (often the x-axis) to the minimum or maximum point

  • E.g. For the graph $y=\mathrm{sin}x$ the amplitude is 1

• The period is the distance along the x-axis to complete a full oscillation

  • Maximum point to maximum point
  • Minimum point to minimum point
  • x-intercept to next but one x-intercept
  • E.g. For the graph $y=\mathrm{cos}x$ the period is 360º

How do I draw an oscillating graph from its equation?

• An oscillating function has equation

  • $\mathrm{f} ( x ) = a \sin ( b x )$
  • $\mathrm{f} ( x ) = a \cos ( b x )$

• The $a$ represents the amplitude of the function

  • The bigger the value of $a$ the bigger the range of values of the function

  • E.g. For the graph $y=5\mathrm{cos}x$

    • The amplitude is 5 units
    • The distance from the x-axis (horizontal line of symmetry) to the maximum/minimum point would be 5 units

  • E.g. For the graph $y=2+\frac{1}{5}\mathrm{cos}x$

    • The amplitude is $\frac{1}{5}$ of a unit
    • The vertical distance from the line $y=2$ to the maximum/minimum point would be $\frac{1}{5}$ of a unit

• The $b$* *determines the period of the function

  • The bigger the value of $b$* *the quicker the function repeats a cycle

  • The period is $\frac{360^{\circ}}{b}$

    • A larger value of $b$ produces a shorter period

  • E.g. For the graph $y = \sin ( 2 x )$

    • The period is $\frac{360}{2}=180^{\circ}$
    • The graph has been 'squashed' by a factor of 2 horizontally
    • Two complete cycles will occur in an interval of $360^{\circ}$

  • E.g. For the graph $y = \sin ( 1 \frac{1}{2} x )$

    • The period is $\frac{360}{\frac{1}{2}}=360\times\frac{2}{1}=720^{\circ}$
    • The graph has been 'stretched'
    • Half a cycle will occur in an interval of $360^{\circ}$

Exam Hint

• You can use your graphic display calculator to help you interpret or compare trigonometric graphs

Worked Example The diagram below shows the graph with equation $y = a \sin ( b x )$, where $a$ and $b$ are integer values.

(a) Write down the period.

The graph intersects the x axis at 0, then again at 45º and again at 90º

One full cycle is complete over an interval of 90º

Final answer: Period = 90º

(b) Write down the amplitude.

The graph goes to a maximum of 3 and a minimum of -3

Final answer: Amplitude = 3

(c) Write down the values of $a$ and $b$.

$a$ is the amplitude

The period is $\frac{360^{\circ}}{b}$* *

$\begin{gathered} 90 = \frac{360}{b} \\ b = \frac{360}{90} \\ b = 4 \end{gathered}$

$a=3\,b=4$

(d) Give a reason as to why the picture cannot be modelled by $y = a \cos ( b x )$.

Final answer: The curve in the diagram goes through the origin$(0,0)$ but a curve with equation of the form$y=a\mathrm{cos}(bx)$ would intercept the y-axis at$(0,a)$

Any suitable reason as to how a cosine graph is different from a sine graph