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Q.
A progressive wave is represented by $y = 12\, sin (5t - 4x) \,cm$. On this wave, how far away are the two points having phase difference of $90^°$ ?
Given equation for the wave is $y =12 \sin (5 t -4 x )$
Phase is given by $5 t -4 x$.
At a given instant, let two points $x _{1}$ and $x _{2}$ differ in phase by $90^{\circ}=\pi / 2$.
$\Rightarrow \left(5 t -4 x _{1}\right)-\left(5 t -4 x _{2}\right)=\pi / 2$
$\Rightarrow 4\left| x _{2}- x _{1}\right|=\pi / 2$
Thus, distance between the two points is $\pi / 8\, cm$