The sources of sound A and B produce a wave of 350   Hz in same phase. A particle P is vibrating under…

The resultant amplitude of waves at the point $P$is, $A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$Here, $\varphi$is the phase difference and $A_1 \text{and} A_2$are amplitude of sound waves.

For phase difference, $\varphi =\frac{2\mathrm{\pi }}{\lambda }∆x$Here, $∆x$is the path difference and $\lambda$is the wavelength of sound wave.

$\Rightarrow \lambda =\frac{v}{f}=\frac{350 \mathrm{m} {\mathrm{s}}^{-1}}{350 \mathrm{Hz}}=100 \mathrm{cm}$

Path difference, $∆x=AP-BP=25 \mathrm{cm}$

Therefore, $\varphi =\frac{2\mathrm{\pi }}{\lambda }∆x =\frac{2\mathrm{\pi }}{100}\times 25=\frac{\mathrm{\pi }}{2}$

Thus, the resultant amplitude is, 

$$\begin{gathered} A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi } \\ A=\sqrt{0.3^2+0.2^2+2\times 0.3\times 0.2\cos \left(\frac{\mathrm{\pi }}{2}\right)} \\ A=\sqrt{0.3^2+0.2^2}=0.5 \mathrm{mm} \end{gathered}$$