Question

A plane progressive simple harmonic sound wave of angular frequency 680 rad/s moves with speed 340 m/s in the direction which makes an equal angle with each x, y and z-axis. The phase difference \[\left( {{\phi _1} - {\phi _2}} \right)\] between the oscillations of the particle in the medium located at the positions \[\left( {\sqrt 3 ,\sqrt 3 ,\sqrt 3 } \right)\] and \[\left( {2\sqrt 3 ,2\sqrt 3 ,2\sqrt 3 } \right)\] is (assume\[\cos \theta > 0\])
A. 2 radian
B. 3 radian
C. 4 radian
D. 6 radian

Answer 1

Hint:The phase of the particle oscillation in a medium is the position with respect to time and space. The rate of change of phase is equal to the speed of the wave. The phase repeats itself after a regular interval of time called the period of the wave.

Formula used:
\[K = \dfrac{{2\pi }}{\lambda }\]
where K is the magnitude of the poynting vector and \[\lambda \]is the wavelength.
\[v = f\lambda \]
where v is the wave speed, f is the frequency and \[\lambda \] is the wavelength.
$y(r,t)=A \sin(\omega t - \overrightarrow k. \overrightarrow r)$
where \[y\left( {\overrightarrow r ,t} \right)\] is the vertical displacement of the particle at position \[\overrightarrow r \] in space and time t. A is the amplitude of the displacement and \[\overrightarrow k \] is the poynting vector.

Complete step by step solution:
It is given that the progressive simple harmonic sound wave is moving in the direction which makes an equal angle with each x, y and z-axis. So, the unit vector along the direction of motion of the progressive simple harmonic sound wave is the unit vector along the poynting vector.
\[\widehat k = \dfrac{{\widehat i + \widehat j + \widehat k}}{{\sqrt 3 }}\]

The angular frequency is given as 680 rad/s per second. So the frequency of the wave is \[f = \dfrac{\omega }{{2\pi }}\]
The magnitude of the Poynting vector is given as,
\[K = \dfrac{{2\pi }}{\lambda }\]
here \[\lambda \] is the wavelength.

Using the relation of wave characteristics,
\[v = f\lambda \\ \]
\[\Rightarrow \lambda = \dfrac{v}{f} \\ \]
\[\Rightarrow \lambda = \dfrac{{340}}{{\left( {\dfrac{{680}}{{2\pi }}} \right)}} \\ \]
\[\Rightarrow \dfrac{\lambda }{{2\pi }} = \dfrac{{340}}{{680}} \\ \]
\[\Rightarrow \dfrac{\lambda }{{2\pi }} = \dfrac{{680}}{{340}} = 2 \\ \]
So, \[K = 2{m^{ - 1}}\]

Hence, the phase difference at two different position will be,
\[{\phi _1} - {\phi _2} = \overrightarrow k \cdot \left( {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} \right) \\ \]
\[\Rightarrow {\phi _1} - {\phi _2} = \dfrac{{2\left( {\widehat i + \widehat j + \widehat k} \right)}}{{\sqrt 3 }} \cdot \left( {\left( {2\sqrt 3 \widehat i + 2\sqrt 3 \widehat j + 2\sqrt 3 \widehat k} \right) - \left( {2\sqrt 3 \widehat i + 2\sqrt 3 \widehat j + 2\sqrt 3 \widehat k} \right)} \right) \\ \]
\[\Rightarrow {\phi _1} - {\phi _2} = \dfrac{{2\sqrt 3 \left( {\widehat i + \widehat j + \widehat k} \right) \cdot \left( {\widehat i + \widehat j + \widehat k} \right)}}{{\sqrt 3 }} \\ \]
\[\therefore {\phi _1} - {\phi _2} = 2 \cdot 3 = 6\] radians
Hence, the phase difference is 6 radians.

Therefore, the correct option is D.

Note: Here it is asked about the phase difference, so we have to find the magnitude of the phase difference. If it is asked as the relative phase of two locations in the space then we have to find the difference with the sign.