Question

How do you find the period and amplitude of $y = 2\sin 3x$?

Answer 1

In this question we have to find the period and amplitude of the given function, we will use the general equation of the $\sin x$ which is given by $y = a\sin \left( {bx + c} \right) + d$, where $a$ is the amplitude, $c$ is the horizontal and $d$ is the vertical shift, and period of the function is given by the formula $\dfrac{{2\pi }}{{\left| b \right|}}$, now substituting the values of the given function we will get the required amplitude and period.

Complete step by step solution:
Now given function is $y = 2\sin 3x$,
Using the general equation of the $\sin x$ which is given by $y = a\sin \left( {bx + c} \right) + d$, where $a$ is the amplitude, $c$ is the horizontal shift and $d$ is the vertical shift, and period of the function is given by the formula $\dfrac{{2\pi }}{{\left| b \right|}}$.
Now rewriting the given function in the standard form we get,
$\Rightarrow y = 2\sin \left( {3x + 0} \right) + 0$,
So here amplitude $a = 2$,$b = 3$,$c = 0$, and$d = 0$,
Now period of the function is given by $\dfrac{{2\pi }}{{\left| b \right|}}$, from the given data substituting the value of $b$ in the formula we get,
Period of the given function will be $\dfrac{{2\pi }}{{\left| 3 \right|}} = \dfrac{{2\pi }}{3}$,
So, the amplitude of the given function is 2 and the period is $\dfrac{{2\pi }}{3}$.
Final Answer:
$\therefore$The amplitude of the given function $y = 2\sin 3x$ is 2 and period will be equal to $\dfrac{{2\pi }}{3}$.

Note:
The graph of $y = \sin x$ is like a wave that forever oscillates between $- 1$ and $1$, in a shape that repeats itself every $2\pi$ units. Specifically, this means that the domain of $\sin x$ is all real numbers, and the range is $\left[ { - 1,1} \right]$.
Properties of $y = \sin x$:
The graph of the function $y = \sin x$ is continuous and extends on either side in symmetrical wave form.
Since the graph intersects the x-axis at the origin and at points where x is an even multiple of ${90^o}$, hence $\sin x$ is zero at$x = n\pi$ where $n = 0, \pm 1, \pm 2, \pm 3.........$.
The ordinate of any point on the graph always lies between 1 and - 1 i.e., $- 1 < y < 1$ or,$- 1 < \sin x < 1$ hence, the maximum value of $\sin x$ is 1 and its minimum value is - 1 and these values occur alternately at$\dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},..........$i. e., at$x = \left( {2n + 1} \right)\dfrac{\pi }{2}$ where $n = 0, \pm 1, \pm 2, \pm 3.........$
Since the function$y = \sin x$ is periodic of period$2\pi$, hence the portion of the graph between $0$ to$2\pi$ is repeated over and over again on either side.