MHT CET · Physics · Waves and Sound
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 \(\left[\cos \left(120^{\circ}\right)=-0.5\right]\)
- A A
- B \(\quad 2 \mathrm{~A}\)
- C \(\frac{\mathrm{A}}{2}\)
- D \(\frac{3 \mathrm{~A}}{2}\)
Answer & Solution
Correct Answer
(A) A
Step-by-step Solution
Detailed explanation
\(\text {Path difference }=\frac{\lambda}{3} \)
\( \therefore \text {Phase difference }=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}=120^{\circ} \)
\( \text {Resultant amplitude } \)
\( \mathrm{R}=\sqrt{\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{~A}_2 \cos \theta} \)
\( \therefore \text {Since, } \mathrm{A}_1=\mathrm{A}_2 \)
\( \therefore \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 120^{\circ}} \)
\( \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \times\left(-\frac{1}{2}\right)} \)
\( \therefore \mathrm{R}=\mathrm{A}\)
\( \therefore \text {Phase difference }=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}=120^{\circ} \)
\( \text {Resultant amplitude } \)
\( \mathrm{R}=\sqrt{\mathrm{A}_1^2+\mathrm{A}_2^2+2 \mathrm{~A}_1 \mathrm{~A}_2 \cos \theta} \)
\( \therefore \text {Since, } \mathrm{A}_1=\mathrm{A}_2 \)
\( \therefore \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 120^{\circ}} \)
\( \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \times\left(-\frac{1}{2}\right)} \)
\( \therefore \mathrm{R}=\mathrm{A}\)
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