Question

Two SHMs are given by\[{{Y}_{1}}=A\sin \left( \dfrac{\pi }{2}t+\phi \right)\] and \[{{Y}_{2}}=B\sin \left( \dfrac{2\pi }{3}t+\phi \right)\]. The phase difference between these two after ‘1’ sec is
 \[A.\,\pi \]
\[B.\,\dfrac{\pi }{2}\]
\[C.\,\dfrac{\pi }{4}\]
\[D.\,\dfrac{\pi }{6}\]

Answer 1

Hint:
The standard equation of SHM is \[y=A\sin (wt+\phi )\], the symbol \[\phi \] represents the phase difference. As we are given with the equations of 2 SHMs, so, we will find the difference of these phase angles and then will substitute the value of time given, in the result obtained.
Formula used:
\[y=A\sin (wt+\phi )\]

Complete step by step answer:
The standard equation of SHM is \[y=A\sin (wt+\phi )\]
Where A is the amplitude of the wave, w is the angular frequency, t is the time taken and \[\phi \]is the phase angle.
The phase difference between the given waves can be found by only subtracting the phase angles of the respective wave equations.
From given, we have the data of the parameters of the waves as follows.
The equation of the first waveform: \[{{Y}_{1}}=A\sin \left( \dfrac{\pi }{2}t+\phi \right)\]
In this, the amplitude of this wave is, A
The angular frequency of this wave is, \[w=\dfrac{\pi }{2}\]
The phase angle is, \[\phi \]
The equation of the second waveform: \[{{Y}_{2}}=B\sin \left( \dfrac{2\pi }{3}t+\phi \right)\]
In this, the amplitude of this wave is, B
The angular frequency of this wave is, \[w=\dfrac{2\pi }{3}\]
The phase angle is, \[\phi \]
Now we will compute the phase difference. So, we have,
\[\begin{align}
  & {{\phi }_{1}}-{{\phi }_{2}}=\dfrac{2\pi t}{3}+\phi -\dfrac{\pi t}{2}-\phi \\
 & \Rightarrow {{\phi }_{1}}-{{\phi }_{2}}=\dfrac{\pi t}{6} \\
\end{align}\]
We are given with the value of time, that is, time = 1 s.
So, substitute the value of the time in the above equation of the phase difference.
\[\begin{align}
  & {{\phi }_{1}}-{{\phi }_{2}}=\dfrac{\pi t}{6} \\
 & \Rightarrow {{\phi }_{1}}-{{\phi }_{2}}=\dfrac{\pi (1)}{6} \\
 & \Rightarrow {{\phi }_{1}}-{{\phi }_{2}}=\dfrac{\pi }{6} \\
\end{align}\]
The phase difference between these two after ‘1’ sec is \[\dfrac{\pi }{6}\].

Thus, the option (D) is correct.

Note:
The phase difference is basically calculated between the phase angles of the waves. But, in this question, we have considered the combination of the angular frequency and the phase angle. Even the units of the parameters should be taken care of. In our case, the time is given in the second unit.