Question

The phase difference between two waves, represented by ${{y}_{1}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+0.5 \right]\,m$ and ${{y}_{2}}={{10}^{-6}}\cos \left[ 100t+\left( {}^{x}/{}_{50} \right) \right]\,m$, where x is expressed in metre and t is expressed in second, is approximately
A. 1.07 radian
B. 2.07 radian
C. 0.5 radian
D. 1.5 radian

Answer 1

The given wave equations should be converted to either of any one form – either in terms of sin function or in terms of cos function. The terms like angular frequency, wave number and the phase of both the equations should be compared to find the total phase difference.

Formula used: $y=A\sin (wt+kx+\phi )$
$\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta$

Complete step by step answer:
From given, we have the data,
The equation of one wave is, ${{y}_{1}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+0.5 \right]\,m$
The equation of the second wave is, ${{y}_{2}}={{10}^{-6}}\cos \left[ 100t+\left( {}^{x}/{}_{50} \right) \right]\,m$
Now convert either of the wave equations in terms of sin function or cos function.
So, we will convert the second wave equation in terms of sin function.
Here, we will use the trigonometric angle properties.
$\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta$
Thus, we have the wave equations as,
${{y}_{1}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+0.5 \right]\,m$
${{y}_{2}}={{10}^{-6}}\sin \left[ \dfrac{\pi }{2}+\left( 100t+\left( {}^{x}/{}_{50} \right) \right) \right]\,m$
Now rearrange the terms within the brackets of the above equation.
${{y}_{1}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+0.5 \right]\,m$ …… (1)
${{y}_{2}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+\dfrac{\pi }{2} \right]\,m$ …… (2)
Find out the values of angular frequencies, wavenumbers and the phase angles.
The angular frequencies are: ${{w}_{1}}=100,{{w}_{2}}=100$
The wave numbers are: ${{k}_{1}}=\dfrac{1}{50},{{k}_{2}}=\dfrac{1}{50}$
The phase angles are: ${{\phi }_{1}}=0.5,{{\phi }_{2}}=\dfrac{\pi }{2}$
Now compare the equations (1) and (2) to find the phase difference between the given waves.
Firstly, compare the angular frequency, the coefficients of the variable ‘t’.

& {{w}_{1}}=100 \\\ & \Rightarrow {{w}_{2}}=100 \\\ & \Rightarrow {{w}_{1}}={{w}_{2}} \\\ \end{aligned}$$ Secondly, compare the wave numbers, the coefficients of the variable ‘x’. $$\begin{aligned} & {{k}_{1}}=\dfrac{1}{50} \\\ & \Rightarrow {{k}_{2}}=\dfrac{1}{50} \\\ & \Rightarrow {{k}_{1}}={{k}_{2}} \\\ \end{aligned}$$ Finally, compare the phase angle values. $$\begin{aligned} & {{\phi }_{1}}=0.5 \\\ & \Rightarrow {{\phi }_{2}}=\dfrac{\pi }{2} \\\ \end{aligned}$$ Thus, we have noticed that there is a difference in phase angles. Therefore, the phase difference is calculated as follows: $$\begin{aligned} & \Delta \phi =\left| {{\phi }_{1}}-{{\phi }_{2}} \right| \\\ & \Rightarrow \Delta \phi =\left| 0.5-\dfrac{\pi }{2} \right| \\\ & \Rightarrow \Delta \phi =\left| 0.5-1.57 \right| \\\ & \Rightarrow \Delta \phi =1.07\,rad \\\ \end{aligned}$$ Therefore, the phase difference is 1.07 radians. As the phase difference between two waves, represented by $${{y}_{1}}={{10}^{-6}}\sin \left[ 100t+\left( {}^{x}/{}_{50} \right)+0.5 \right]\,m$$ and $${{y}_{2}}={{10}^{-6}}\cos \left[ 100t+\left( {}^{x}/{}_{50} \right) \right]\,m$$ is approximately 1.07 radians **So, the correct answer is “Option A”.** **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem. The uniformity in the case of units of the parameters should be maintained. Without converting to either of sin or cos functions, solving the problem will be the wrong method.