Question

How do you find the amplitude and period of $y = 2\sin \left( {\pi .x} \right)$ ?

Answer 1

Hint : We have been given a sinusoidal equation or function to find the amplitude and the period. Such functions are also known as wave functions. A wave function means that the given function forms a continuous and repetitive wave pattern. The general form of a wave function is given as,
$y = A\sin \left( {Bx + C} \right) + D$
where, $A$ is the amplitude, $B$ denotes time period, $C$ is the phase shift and $D$ is the vertical shift of the equation.
The amplitude is the maximum vertical stretch of the wave at crest and trough, and period is the time after which the waves are being repeated.
We can compare the given equation with the general form to find the amplitude and period.

Complete step by step solution:
We have been given the function $y = 2\sin \left( {\pi .x} \right)$ .
We will compare it with the general form of the wave function,
$y = A\sin \left( {Bx + C} \right) + D$
First we write the given equation in general form as,
$y = 2\sin \left( {\pi x + 0} \right) + 0$
On comparison we can observe that,
$A = 2$ , $B = \pi$ , $C = 0$ and $D = 0$ .
Amplitude of the function is given as $\left| A \right|$ .
Thus, $Amplitude = \left| A \right| = \left| 2 \right| = 2$
Period of the function is given as $\dfrac{{2\pi }}{{\left| B \right|}}$ .
Thus, $Period = \dfrac{{2\pi }}{{\left| B \right|}} = \dfrac{{2\pi }}{{\left| \pi \right|}} = \dfrac{{2\pi }}{\pi } = 2$
Hence, the amplitude of the given function is $2$ and period of the given function is $2$ .
So, the correct answer is “2”.

Note : For finding the amplitude and period we compared the given equation with the general form. We used absolute values as amplitude and period are both positive terms and cannot be negative. An amplitude of $2$ means that the wave of the given function extends to a maximum of $2\;units$ at crest and trough. A period of $2\;units$ means that the wave gets repeated after a time period of $2\;units$ .