Question

The phase difference between two waves represented by
${y_1} = {\text{sin}}\left[ {100t + 50x + 0.5} \right]m$
${y_2} = \cos \left[ {100t + 50x} \right]m$
where x is expressed in metres and t is expressed in seconds is approximately
A. 1.07 rad
B. 2.01 rad
C. 0.5 rad
D. 1.5 rad

Answer 1

Phase of the wave represents the location of the point within a specific wave form and generally that location varies with respect to time and position too. Now if we want to view a wave we can do it by taking a snap that is by putting time as a fixed one and varying position or we can either fix a position and get the shape of the wave which varies with time.
Formula used:
\eqalign{ & Y = A\sin ( \pm wt \pm kx) \cr & Y = A\cos ( \pm wt \pm kx) \cr}

Complete answer:
Usually most of the wave equations we frequently use will be of forms
\eqalign{ & Y = A\sin ( \pm wt \pm kx) \cr & Y = A\cos ( \pm wt \pm kx) \cr}
Where $w$ is the angular frequency and $k$ is the angular number.
The part which is present inside the sin or cos functions is called the phase of the wave and if we want to find the phase difference of the waves we have to subtract one phase from the other to get it.
We were given with two wave functions
${y_1} = {\text{sin}}\left[ {100t + 50x + 0.5} \right]m,{y_2} = \cos \left[ {100t + 50x} \right]m$
In order to find out the phase difference between two waveforms we have to convert both of them into either sinusoidal function or cosecant function.
Let us convert the second function into the sinusoidal function
We know
$\sin (\dfrac{\pi }{2} + \theta ) = \cos (\theta )$
Hence ${y_2} = \cos \left[ {100t + 50x} \right]m = \sin (\dfrac{\pi }{2} + 100t + 50x)m$
Now phase difference will be $[(\dfrac{\pi }{2} + 100t + 50x) - (100t + 50x + 0.5)] = [\dfrac{\pi }{2} - 0.5] = 1.07{\text{rad}}$

So, the correct answer is “Option A”.

Note:
In this case we got constant phase difference which didn’t vary with position or time but sometimes we get phase difference which varies with only time or only position or with both time and position. If phase difference is constant then fringe pattern can’t be formed the equations which we use to denote various S.H.M’s will have constant phase difference hence fringes can’t be formed here.