Two identical travelling waves, moving in the same direction, are out of phase by \[\dfrac{\pi }{2}rad\]. What is the amplitude of the resultant wave in terms of the common amplitude \[{{y}_{m}}\] of the two waves.

Hint: The formula that is used to compute the amplitude of the resultant wave in terms of the amplitudes of waves and the cosine of phase angle between them should be used to solve the problem. As the amplitudes of the waves and the phase difference is given, so, substituting the same, we will compute the value.
Formula used:
\[R=\sqrt{A_{1}^{2}+A_{2}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi }\]

Complete step by step answer:
From the given information, we have the data as follows.
Two identical travelling waves, moving in the same direction, are out of phase by \[\dfrac{\pi }{2}rad\].
The amplitudes of 2 waves are equal, represented as,\[{{y}_{m}}\].
The phase difference between these 2 waves is given as,\[\phi =\dfrac{\pi }{2}\].
The resultant wave of the waves is given by the formula as follows.
\[R=\sqrt{A_{1}^{2}+A_{2}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi }\]
Where R is the amplitude of the resultant, \[{{A}_{1}}\]is the amplitude of the first wave, \[{{A}_{2}}\]is the amplitude of the second wave, \[\phi \]is the phase difference between the waves.
From the given information, we have the data as follows.
The amplitude of the first wave is, \[{{A}_{1}}={{y}_{m}}\]
The amplitude of the second wave is, \[{{A}_{2}}={{y}_{m}}\]
The phase difference between these waves is, \[\phi =\dfrac{\pi }{2}\]
Substitute the values of the given parameters in the above equation.
\[\begin{align}
  & R=\sqrt{y_{m}^{2}+y_{m}^{2}+2\times {{y}_{m}}\times {{y}_{m}}\times \cos \dfrac{\pi }{2}} \\
 & \Rightarrow R=\sqrt{2y_{m}^{2}+2y_{m}^{2}\times 0} \\
 & \Rightarrow R=\sqrt{2y_{m}^{2}} \\
 & \therefore R=\sqrt{2}{{y}_{m}} \\
\end{align}\]

\[\therefore \] The amplitude of the resultant wave in terms of the amplitude of the waves is, \[\sqrt{2}{{y}_{m}}\].

Note: In this case, the amplitude of the out of phase waves is given to be the same. But, in the case of the out of waves, one of the waves will have the same frequency shifted by one-half cycle concerning the frequency of the other wave, and in turn, there will be a difference in their amplitudes.