Question

Two sources of sound A and B produce the wave of $350Hz$, they vibrate in the same phase. The particle $P$ is vibrating under the influence of these two waves. If the amplitude at the point $P$ produced by the two waves is $0.3mm$ and $0.4mm$ then the resultant amplitude of the point $P$ will be: (path difference $AP-BP=25cm$ and the velocity of sound is $350m/sec$)

• A. $0.7mm$
• B. $0.1mm$
• C. $0.2mm$
• D. $0.5mm$

Answer 1

The correct option is D $0.5mm$

$\begin{array}{c}\lambda =\frac{v}{n}=\frac{350}{350}=1m=100cm\\ Pathdifference\Delta x=AP-BP=25cm\\ Phasedifference\Delta \phi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{1}\times \left(\frac{25}{100}\right)=\frac{\pi }{2}\\ AmplitudeA=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2}=\sqrt{{\left(0.3\right)}^2+{\left(0.4\right)}^2}=0.5mm\end{array}$