Question

Two sources of sound $A$ and $B$ produces the wave of $350\, Hz$, they vibrate in the same phase. The particle $P$ is vibrating under the influence of these two waves, if the amplitudes at the point $P$ produced by the two waves is $0.3\, mm$ and $0.4\, mm$, then the resultant amplitude of the point $P$ will be when $A P-B P=25\, cm$ and the velocity of sound is $350\, m / \sec$

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

Answer 1

Answer: (D)

$\lambda=\frac{v}{n}=\frac{350}{350}=1\, m =100\, cm$
Also path difference $(\Delta x)$ between the waves at the point of observation is $A P-B P=25\, cm$.
Hence $\Rightarrow \Delta \varphi=\frac{2 \pi}{\lambda}(\Delta x)=\frac{2 \pi}{1} \times\left(\frac{25}{100}\right)=\frac{\pi}{2}$
$\Rightarrow A=\sqrt{\left(a_{1}\right)^{2}+\left(a_{2}\right)^{2}}$
$=\sqrt{(0.3)^{2}+(0.4)^{2}}=0.5\, mm$