The equation of progressive wave is $Y = 4\sin \left\{ {\pi \left( {\dfrac{t}{5} - \dfrac{x}{9}} \right) + \dfrac{\pi }{6}} \right\}$ where x and y are in cm. Which of the following statements is true?
$
 {\text{A)}}\lambda {\text{ = 18cm}} \\
  {\text{B) Amplitude = 0}}{\text{.04cm}} \\
  {\text{C) velocity, v = 50cm/s}} \\
  {\text{D) frequency, f = 20Hz}} \\
 $

Hint: The term progressive wave equation is defined as when the two expressions of the wave function of the time and space are joined then it is known as progressive wave equation.
The standard equation of the progressive wave is
$Y = A\sin \left[ {2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right) + \phi } \right]$

Complete step by step solution:
From the standard wave equation
$Y = A\sin \left[ {2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right) + \phi } \right]$
Here Y is the function to find the amplitude, time period and frequency.
A is the amplitude
T is the Time period
$\lambda $ is the wavelength
$\phi $ is the phase difference
Now, we compare the standard wave equation and give progressive wave equation.
 $Y = 4\sin \left\{ {\pi \left( {\dfrac{t}{5} - \dfrac{x}{9}} \right) + \dfrac{\pi }{6}} \right\}$
Here, we just change a few things for comparison purposes.
$Y = 4\sin \left\{ {2\pi \left( {\dfrac{t}{{10}} - \dfrac{x}{{18}}} \right) + \dfrac{\pi }{6}} \right\}$---(1)
$Y = A\sin \left[ {2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right) + \phi } \right]$ --(2)
From comparison we get the values of amplitude, wavelength, and time period.
Amplitude, A = 4cm
Time Period, T=10s
Wavelength, $\lambda $ =18cm
Phase difference, $\phi $ = $\dfrac{\pi}{6}$
Hence, the option (A) is correct.

Note: The progressive wave distributed by the point source to the surrounding area. Here’s an another wave equation
$Y(x,t) = A\sin (kx - \omega t + \phi )$
Here we should know the terms of the progressive wave equation otherwise a small confusion can make the solution incorrect.