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 AP – BP = 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)

0.5 mm

Answer 2

$\lambda = \frac{v}{n} = \frac{350}{350} = 1m$=100 cm

Also path difference $(\Delta x)$ between the waves at the point of observation is $AP - BP = 25cm$. Hence

⇒ $\Delta\varphi = \frac{2\pi}{\lambda}(\Delta x) = \frac{2\pi}{1} \times \left( \frac{25}{100} \right) = \frac{\pi}{2}$

⇒ $A = \sqrt{(a_{1})^{2} + (a_{2})^{2}}$ $=$ $\sqrt{(0.3)^{2} + (0.4)^{2}}$= 0.5 $mm$