The pressure variation in a sound wave in air is given by $\mathrm{\Delta p}=12\sin \left(8.18\mathrm{x}-2700\mathrm{t}+\frac{\mathrm{\pi }}{4}\right){\mathrm{Nm}}^{-2}$ where, x and t are in SI units. If density of air is 1.29 kgm^-3, the displacement amplitude is

From the given equation , $(\mathrm{\Delta p}{)}_{\mathrm{m}}=12\mathrm{Pa}$
$\begin{array}{l}\mathrm{k}=8.18{\mathrm{m}}^{-1}\text{ and }\mathrm{\omega }=2700 \mathrm{rad} {\mathrm{s}}^{-1}\\ \Rightarrow \mathrm{\lambda }=\frac{2\mathrm{\pi }}{\mathrm{k}}=\frac{2\mathrm{\pi }}{8.18} \text{ and }\mathrm{f}=\frac{\mathrm{\omega }}{2\mathrm{\pi }}=\frac{2700}{2\mathrm{\pi }} {\mathrm{s}}^{-1}\\ \Rightarrow \mathrm{v}=\mathrm{f\lambda }=\frac{2700}{2\mathrm{\pi }}\times \frac{2\mathrm{\pi }}{8.18}=330 {\mathrm{ms}}^{-1}\\ \mathrm{v}=\sqrt{\frac{\mathrm{B}}{\mathrm{\rho }}}\\ \Rightarrow \mathrm{B}={\mathrm{\rho v}}^2=1.29\times (330{)}^2=140.5\times {10}^3 {\mathrm{Nm}}^{-2}\end{array}$
We know $(\mathrm{\Delta p}{)}_{\mathrm{m}}=\mathrm{BAk}$
$\Rightarrow \mathrm{A}=\frac{(\mathrm{\Delta p}{)}_{\mathrm{m}}}{\mathrm{Bk}}=\frac{12}{140.5\times {10}^3\times 8.18}=1.044\times {10}^{-5}\mathrm{m}$