Question

Two waves are represented by the equations
${y_1} = a\sin \left( {\omega t + kx + 0.57} \right)m$ and
${y_2} = a\cos \left( {\omega t + kx} \right)m$
Where $x$ is in $meter$ and $t$ in $\sec$ . The phase difference between them is
A. $0.57\,radian$
B. $1.0\,radian$
C. $1.25\,radian$
D. $1.57\,radian$

Answer 1

Here, we will compare the first equation with ${y_1} = a\sin {\phi _1}$ and the second equation with ${y_2} = a\sin {\phi _2}$ . From this we will get the values of the values of phases ${\phi _1}$ and ${\phi _2}$ . The phase difference between them will be calculated by subtracting these phases. As a result, we will get the phase difference between the two waves.

Complete step by step answer:
The waves given in the above question are in the form of a displacement of transverse harmonic wave that are travelling in the negative X-direction.
Now, as given in the question, the displacement ${y_1}$ of the first equation is given by
${y_1} = a\sin \left( {\omega t + kx + 0.57} \right)m$
Here $x$ will be in $meter$ and $t$ will be in $\sec$
Comparing, the above equation with the equation ${y_1} = a\sin {\phi _1}$ , we get the phase of the first wave
${\phi _1} = \left( {\omega t + kx + 0.57} \right)m$
Also, the displacement ${y_2}$ of the second wave is given by
${y_2} = a\cos \left( {\omega t + kx} \right)m$
Now, the above equation can be written in terms of $\sin$ as shown below
${y_2} = a\sin \left( {\dfrac{\pi }{2} + \omega t + kx} \right)m$
$\Rightarrow \,{y_2} = a\sin \left( {\omega t + kx + \dfrac{\pi }{2}} \right)m$
Now, comparing the above equation with the equation ${y_2} = a\sin {\phi _2}$ , we get the phase of the second wave
${\phi _2} = \omega t + kx + \dfrac{\pi }{2}$
Now, the phase difference between the two waves can be calculating by subtracting both the two phases and is given by
$\Delta \phi = {\phi _2} - {\phi _1}$
Putting the values of ${\phi _1}$ and ${\phi _2}$ in the above equation, we get
$\Delta \phi = \left( {\omega t + kx + \dfrac{\pi }{2}} \right)m - \left( {\omega t + kx + 0.57} \right)m$
$\Rightarrow \,\Delta \phi = \left( {\dfrac{\pi }{2} - 0.57} \right)m$
Now, the value of $\dfrac{\pi }{2} = 1.57$
Putting this value of $\dfrac{\pi }{2}$ in the above equation, we get
$\Delta \phi = \left( {1.57 - 0.57} \right)m$
$\therefore \,\Delta \phi = 1.0$
Therefore, the phase difference between the two waves is $1.0\,radian$ .

Hence, option B is the correct option.

Note: Here, in the above solution, we have the value of $\dfrac{\pi }{2} = 1.57$ . you might get confused about how this value comes. This value of $\dfrac{\pi }{2}$ can be calculated as shown below
$\dfrac{\pi }{2} = \dfrac{{3.14}}{2}$
$\Rightarrow \,\dfrac{\pi }{2} = 1.57$
This is the required value.