A transverse harmonic wave on a string is described by $y(x,t) = 3.0\sin (36t + 0.018x + \dfrac{\pi }{4})$ where $x$ and $y$ are in $cm$ and $t$ in $s$. The position of $x$ is from left to right.
A. Is this a travelling wave or a stationary wave?
If it is a travelling wave what is the speed and direction of its propagation?
B. What is its amplitude and frequency?
C. What is the initial phase at the origin?
D. What is the least distance between two successive crests in the wave?

Hint: Here we have to first find the equation given by the displacement function and compare the given equation and the displacement equation to get the speed. Vibration of a wave is represented by a transverse harmonic wave.
Then we have to find the frequency of the given wave by using the formula between frequency and angular velocity.
Also we can get the phase angle by comparing the displacement equation and the given equation.
Lastly, we can find the wavelength by determining the distance between the successive crests.

Complete step by step solution:
(A). The displacement equation of wave travelling from left to right is given by:
$y(x,t) = a\sin (\omega t + kx + \phi )$
               ...... (1)
The given equation in the question is:
$y(x,t) = 3.0\sin (36t + 0.018x + \dfrac{\pi }{4})$
               ...... (2)
When we compare the two equations, we find that equation (2) also represents the displacement equation of a wave. So, it is a travelling wave.
From equations (1) and (2) we get,
$\omega = 36\,rad/s$, $k = 0.018\,{m^{ - 1}}$

We know that,
Speed, $v = \dfrac{\omega }{{2\pi }}$ and wavelength, $\lambda = \dfrac{{2\pi }}{k}$

Also,
$
 v = f\lambda \\
 \therefore v = \dfrac{\omega }{{2\pi }} \times \dfrac{{2\pi }}{k} = \dfrac{\omega }{k} \\
 v = \dfrac{{36}}{{0.018}} = 2000\,cm/s = 20\,m/s \\
$

Hence, the speed of the given travelling wave is $20\,m/s$

(B). From the equations (1) and (2), we find that
Amplitude of the given wave, $a = 3\,cm$

Frequency of the given wave, $f = \dfrac{\omega }{{2\pi }} = \dfrac{{36}}{2} \times 3.14 = 573\,Hz$

(C). Again comparing equations (1) and (2), the initial phase angle, $\phi = \dfrac{\pi }{4}$

(D). The least distance between two successive crests in the wave is equal to the wavelength of the wave.
Wavelength, $\lambda = \dfrac{{2\pi }}{k} = \dfrac{{2 \times 3.14}}{{0.018}} = 348.89\,cm = 3.49\,m$

Note: Here we have to remember only the displacement equations otherwise the simplification of this question is not possible. Also since the wave has both frequency and wavelength, we can assume the wave as a travelling wave and not a stationary one.