Correct option is D $72^{\circ}$
Given equation can be written as $y=6 \sin (4 \pi t-0.2 \pi x)$
$\therefore$ Phase $\phi=-0.2 \pi x$
$\Rightarrow \phi_{1}=-0.2 \pi \mathrm{x}_{1}$ and $\phi_{2}=-0.2 \pi \times 2$
$\therefore$ Phase difference $\Delta \phi=\left|\phi_{2}-\phi_{1}\right|=\left|-0.2 \pi\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)\right|$
given $\left(x_{2}-x_{1}\right)=2 \mathrm{~mm}$ substituting in above equation we get $\Delta \phi=0.2 \times$ $180^{0} \times 2=72^{0}$
The equation of simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $m m$ and $t$ is in seconds . The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is
A $18^{0}$
B $36^{\circ}$
C $54^{0}$
D $72^{0}$
A $18^{0}$
B $36^{\circ}$
C $54^{0}$
D $72^{0}$
The equation of simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0.1 x)$, where $x$ and $y$ are in $m m$ and $t$ is in seconds . The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is
A $18^{0}$
B $36^{\circ}$
C $54^{0}$
D $72^{0}$
A $18^{0}$
B $36^{\circ}$
C $54^{0}$
D $72^{0}$