influence of these two waves, if the amplitudes at the point $P$ produced by the two waves are $0.3\text{…
influence of these two waves, if the amplitudes at the point $P$ produced by the two waves are $0.3\text{ mm}$ and $0.4\text{ mm}$, then the resultant amplitude of the point $P$, when $AP - BP = 25\text{ cm}$ and the velocity of sound is $350\text{ ms}^{-1}$, will be
0.7 mm
0.1 mm
0.2 mm
0.5 mm
Solution
Wavelength, $\lambda = \frac{v}{n} = \frac{350}{350} = 1\text{ m} = 100\text{ cm}$
Also, path difference $(\Delta x)$ between the waves at the point of observation is $AP - BP = 25\text{ cm}$. Hence,
Phase difference, $\Delta\phi = \frac{2\pi}{\lambda} (\Delta x) = \frac{2\pi}{1} \times \left(\frac{25}{100}\right) = \frac{\pi}{2}$
$\Rightarrow$ Resultant amplitude, $A = \sqrt{(a_1)^2 + (a_2)^2}$
$= \sqrt{(0.3)^2 + (0.4)^2} = 0.5\text{ mm}$