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 $350 m/sec$ .... $mm$
Diffcult
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(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 $AP - BP = 25cm$.
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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