MCQ
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 $350 m/sec$  .... $mm$
  • A
    $0.7$
  • B
    $0.1$
  • C
    $0.2$
  • $0.5$

Answer

Correct option: D.
$0.5$
d
(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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