Find the distance between a node and next antinodes for the stationary wave if $y = 4\sin \left( {\dfrac{{\pi x}}{{15}}} \right)\cos \left( {96\pi t} \right)$.
A) $7.5$
B) $15$
C) $22.5$
D) $30$

Hint: Find the general equation for the stationary wave then compare it with the equation from the question and find the value of wavelength. Now, use the relation between distance and wavelength.

Complete step by step answer:
Let us suppose that a wave is travelling in the direction of $x - axis$ having uniform velocity $v$, angular frequency $\omega $with amplitude $a$ and wavelength $\lambda $ and wave number is $k$.
The displacement of particle at $x - axis$ in time $t$ is –
${y_1} = a\sin \left( {kx - \omega t} \right)$
Let there be another wave with same amplitude, velocity, frequency, wavelength and wave number but in the negative direction of $x - axis$ because of deflection –
$\Rightarrow$ ${y_2} = a\sin \left( {kx + \omega t} \right)$
Now, using the superposition principle, the resultant displacement of the waves is –
$\Rightarrow$ $y = {y_1} + {y_2} \cdots \left( 1 \right)$
Putting the values of two waves in the equation $\left( 1 \right)$, we get –
$ \Rightarrow y = a\left[ {\sin \left( {kx + \omega t} \right) + \sin \left( {kx - \omega t} \right)} \right] \cdots \left( 2 \right)$
Now, we know that, -
$\sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B$
So, the equation $\left( 2 \right)$, can be written as –
$\Rightarrow$ $y = 2a\sin \left( {kx} \right)\cos \left( {\omega t} \right) \cdots \left( 3 \right)$
The effect of two equal waves in opposite directions is that the amplitude at each value of x is fixed in time.  Amplitude of the vibrating particles in the wave $ = 0$ if $kx = n\pi $, where $n$ is an integer. Amplitude becomes maximum when $kx = \left( {n + \dfrac{1}{2}} \right)\pi $
Stationary waves are those in which the elements in the medium oscillate with the same phase. They have fixed positions in space and move synchronously.
Nodes are the points where amplitude is zero, that is, $kx = n\pi $
We know that, $k = \dfrac{{2\pi }}{\lambda }$
$\therefore x = \dfrac{{n\lambda }}{2}$ where, $n = 0,1,2,3,4,5$
Antinodes are the points where amplitude of oscillation is maximum, that is, $kx = \left( {n + \dfrac{1}{2}} \right)\pi $
$\therefore x = \left( {n + \dfrac{1}{2}} \right)\dfrac{\lambda }{2}$ where, $n = 0,1,2,3,4$
Let the second wave is a wave reflected from an obstruction that the progressive wave encounters on the medium.  It has a phase difference of $\pi $.
So, the equation of stationary wave for nodes and anti – nodes can be written as –
$\Rightarrow$ $y = 2a\sin \dfrac{{2\pi x}}{\lambda }\cos \dfrac{{2\pi vt}}{\lambda } \cdots \left( 3 \right)$
According to the question, the equation of given wave is –
$\Rightarrow$ $y = 4\sin \left( {\dfrac{{\pi x}}{{15}}} \right)\cos \left( {96\pi t} \right) \cdots \left( 4 \right)$
Comparing equation $\left( 3 \right)\& \left( 4 \right)$, we get –
$
  \dfrac{{2\pi }}{\lambda } = \dfrac{\pi }{{15}} \\
   \Rightarrow \lambda = 2 \times 15 \\
   \Rightarrow \lambda = 30 \\
 $
Now, the distance between the nearest node and antinode can be measured as –
$
   \Rightarrow \dfrac{\lambda }{4} \\
   \Rightarrow \dfrac{{30}}{4} = 7.5 \\
 $
Hence, the distance between a node and next antinodes for the given stationary wave is $7.5$.
So, the correct option is (A).

Note: The combination of the interference of the two waves is called standing wave pattern. Standing waves are produced when they are travelling in the opposite direction within the same medium and the waves have the same frequency and amplitude.