Question

The equation of a stationary wave is $y=2\sin \left( \dfrac{\pi x}{15} \right)\cos \left( 48\pi t \right)$ . The distance between a node and its next antinode is
A. $7.5units$
B. $22.5units$
C. $1.5units$
D. $30units$

Answer 1

Hint:
To solve the given question we need to dive into the concept of wave number and nodes.
Wave number refers to the number of complete wave cycles of an electromagnetic field that exist in one meter of linear space. It is generally expressed as $k=\dfrac{2\pi }{\lambda }$ in reciprocal meters $\left( {{m}^{-1}} \right)$
Nodes are basically the points on a wave where the amplitude is minimum or zero and are found in the mean positions of the stationary wave whereas antinodes are the points on a wave where the amplitude is maximum and are found on the crests and troughs of the stationary wave.

Complete answer:
The information given to us is that of a stationary wave is $y=2\sin \left( \dfrac{\pi x}{15} \right)\cos \left( 48\pi t \right)$
And wave number of the given stationary wave, $k=\dfrac{\pi }{15}$
Thus, equating the given value of wavenumber to its general expression,
$\dfrac{2\pi }{\lambda }=\dfrac{\pi }{15}$
$\Rightarrow \lambda =30$
Now, one might notice that the distance between two adjacent equilibrium positions or nodes is $\dfrac{\lambda }{2}$ where as the distance between antinodes which is the distance between one adjacent crest and trough is also $\dfrac{\lambda }{2}$ but the distance between one node to its adjacent antinode is $\dfrac{\lambda }{4}$.
Hence, the required distance between a node and its next antinode $=\dfrac{\lambda }{4}=\dfrac{30}{4}=7.5units$

Therefore, the correct option would be (A) $7.5units$

Note:
One is advised to remember that though wave number or $k$ here is expressed as $\dfrac{2\pi }{\lambda }$ but alternatively it can also be expressed as the ratio of angular velocity to phase or group velocity, which is basically more elaborated method of the already given expression.