wave_mechanics interference_and_superposition_of_waves Question 7
Question: Two waves are propagating to the point P along a straight line produced by two sources A and B of simple harmonic and of equal frequency. The amplitude of every wave at P is a and the phase of A is ahead by $ \frac{\pi }{3} $ than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter
[BVP 2003]
Options:
A) 2a
B) $ a\sqrt{3} $
C) $ a\sqrt{2} $
D) a
Show Answer
Answer:
Correct Answer: D
Solution:
Path difference $ (\Delta x) $
$ =50cm=\frac{1}{2}m $
$ \therefore $ Phase difference $ \Delta \varphi =\frac{2\pi }{\lambda }\times $
$ \Delta x\Rightarrow \varphi =\frac{2\pi }{1}\times \frac{1}{2}=\pi $ Total phase difference = $ \pi -\frac{\pi }{3}=\frac{2\pi }{3} $
Therefore $ A=\sqrt{a^{2}+a^{2}+2a^{2}\cos (2\pi /3)}=a $
BETA
