Q165 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
Two waves having same amplitude $A$, angular frequency $\omega$ and propagation constant $k$, traveling in opposite directions, have a phase difference of $60^\circ$. If they interfere with each other, the resultant displacement $y$ of the wave is
- Correct answerA$2A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
- B$A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
- C$2A\sin(kt + 30^\circ)\cos(\omega x + 30^\circ)$
- D$2A\sin(\omega t + kx + 60^\circ)$
Solution
Let the equations of the two waves traveling in opposite directions be $y_1 = A\sin(\omega t - kx)$ and $y_2 = A\sin(\omega t + kx + \phi)$.
Given that the phase difference is $\phi = 60^\circ$.
The resultant displacement is given by the principle of superposition:
$y = y_1 + y_2$
$y = A\sin(\omega t - kx) + A\sin(\omega t + kx + 60^\circ)$
Using the trigonometric identity $\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$, we get:
$y = 2A\sin\left(\dfrac{\omega t - kx + \omega t + kx + 60^\circ}{2}\right)\cos\left(\dfrac{\omega t + kx + 60^\circ - (\omega t - kx)}{2}\right)$
$y = 2A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
Answer: $2A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
Given that the phase difference is $\phi = 60^\circ$.
The resultant displacement is given by the principle of superposition:
$y = y_1 + y_2$
$y = A\sin(\omega t - kx) + A\sin(\omega t + kx + 60^\circ)$
Using the trigonometric identity $\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$, we get:
$y = 2A\sin\left(\dfrac{\omega t - kx + \omega t + kx + 60^\circ}{2}\right)\cos\left(\dfrac{\omega t + kx + 60^\circ - (\omega t - kx)}{2}\right)$
$y = 2A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
Answer: $2A\sin(\omega t + 30^\circ)\cos(kx + 30^\circ)$
Hence, the correct answer is option: 'A'.