$\Rightarrow \frac{f_{1}}{f_{2}}=\frac{c+v}{c-v} \Rightarrow \frac{100}{90}=\frac{c+v}{c-v}$
$\Rightarrow 100 c-100 v=90 C+90 v$
$\Rightarrow 10 c=190 v$
$\Rightarrow 1 / 19 \times 330=17.3 \mathrm{m} / \mathrm{s}$
${y_1} = 2a\sin (\omega t - kx)$ and ${y_2} = 2a\sin (\omega t - kx - \theta )$
The amplitude of the medium particle will be