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 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]

• A. 2a
• B.
• C.
• D. a

Answer 1

The correct option is D

a

Path difference $\left(△x\right)$ = 50 cm =$\frac{1}{2}m$

$\therefore$ Phase difference $△\varphi =\frac{2\pi }{\lambda }\times △x\Rightarrow \varphi =\frac{2\pi }{1}\times \frac{1}{2}=\pi$

Total phase difference =$\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$\Rightarrow A=\sqrt{a^2+a^2+2a^{2}cos\left(2\pi /3\right)}=a$