Question

Two sources of sound $\text{A}$ and $\text{B}$ produce waves of $350\text{Hz}$. The particle at point $\text{P}$ is vibrating under the influence of these two waves. The amplitudes of the two waves are $0.3\text{mm}$ and $0.4\text{mm}$. If $AP-BP=25\text{cm}$ and velocity of sound is $350\text{m/s}$, the resultant amplitude of the particle at point $\text{P}$ will be

• A. $0.7\text{mm}$
• B. $0.1\text{mm}$
• C. $0.2\text{mm}$
• D. $0.5\text{mm}$

Answer 1

The correct option is D $0.5\text{mm}$

We know that ,
Wavelength, $\lambda =\frac{v}{f}=\frac{350}{350}=1\text{m}\text{or}100\text{cm}$

Also, path difference $\left(\Delta x\right)$ between the waves at the point of observation is $AP-BP=25\text{cm}$.
Hence, phase difference $\Delta \varphi =\frac{2\pi }{\lambda }\times \left(\Delta x\right)=\frac{2\pi }{100}\times 25=\frac{\pi }{2}$

$\therefore$ Amplitude of resultant wave
$A=\sqrt{a_1^2+a_2^2}$
From the data given in the question,
$A=\sqrt{{\left(0.3\right)}^2+{\left(0.4\right)}^2}=0.5\text{mm}$

Hence, option (d) is the correct answer.