Question

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

Answer 1

Hint: 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.