Two sound waves each of wavelength ' $\lambda$ ' and having the same amplitude ' A ' from two source '…

Two sound waves each of wavelength ' $\lambda$ ' and having the same amplitude ' A ' from two source ' $\mathrm{S}_1$ ' and ' $\mathrm{S}_2$ ' interfere at a point P . If the path difference, $\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}=\lambda / 3$ then the amplitude of resultant wave at point ' P ' will be $\left[\cos \left(120^{\circ}\right)=-0.5\right]$
  1. A
  2. $\quad 2 \mathrm{~A}$
  3. $\frac{\mathrm{A}}{2}$
  4. $\frac{3 \mathrm{~A}}{2}$

Solution

$\begin{array}{ll} & \text { Path difference }=\frac{\lambda}{3} \\ \therefore \quad & \text { Phase difference }=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}=120^{\circ} \\ & \text { Resultant amplitude } \\ & \mathrm{R}=\sqrt{\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{~A}_2 \cos \theta} \\ \therefore \quad & \text { Since, } \mathrm{A}_1=\mathrm{A}_2 \\ \therefore \quad & \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 120^{\circ}} \\ \quad & \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \times\left(-\frac{1}{2}\right)} \\ \therefore \quad & \mathrm{R}=\mathrm{A}\end{array}$

Asked in: MHT CET 2024 (16 May Shift 1)

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