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
Text Solution
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$\lambda = \frac{v}{n} = \frac{350}{350} = 1m = 100 \text{ cm}$
Also path difference $(\Delta x)$ between the waves at the point of observation is $AP - BP = 25cm$ . Hence
⇒ ⇒ $\Delta \phi = \frac{2 \pi}{\lambda} (\Delta x) = \frac{2 \pi}{1} \times \left(\frac{25}{100}\right) = \frac{\pi}{2}$
⇒ ⇒ $A = \sqrt{(a_1)^2 + (a_2)^2}$ ___ $\sqrt{(0.3)^2 + (0.4)^2}$ = 0.5 mm
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