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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

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0.7 mm

0.1 mm

0.2 mm

0.5 mm

answer is D.

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Detailed Solution

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

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

$\Rightarrow ∆ φ = \frac{2 π}{λ} (∆ x) = \frac{2 π}{1} \times (\frac{25}{100}) = \frac{π}{2}$

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

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