JEE Physics 2025 Question with Solution

Given:

• $y_1(x, t) = 4\sin(kx - \omega t)$
• $y_2(x, t) = 2\sin\left(kx - \omega t + \frac{2\pi}{3}\right)$

Find: The amplitude and phase of the resultant wave.

For the superposition of two sinusoidal waves of the same angular frequency, write the resultant as

$$y(x, t) = A\sin(kx - \omega t + \phi)$$

The resultant amplitude is

$$A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos\delta}$$

Here,

• $a_1 = 4$
• $a_2 = 2$
• $\delta = \frac{2\pi}{3}$

Substituting,

$$A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos\left(\frac{2\pi}{3}\right)}$$

Since

$$\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$

we get

$$A = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3}$$

Now for the phase,

$$\tan\phi = \frac{a_2\sin\delta}{a_1 + a_2\cos\delta}$$

Substituting the values,

$$\tan\phi = \frac{2\sin\left(\frac{2\pi}{3}\right)}{4 + 2\cos\left(\frac{2\pi}{3}\right)}$$

Using

$$\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}, \qquad \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$

we get

$$\tan\phi = \frac{2 \times \frac{\sqrt{3}}{2}}{4 - 1} = \frac{\sqrt{3}}{3}$$

Hence,

$$\phi = \frac{\pi}{6}$$

Therefore, the amplitude and phase of the resultant wave are $[2\sqrt{3}, \frac{\pi}{6}]$. The correct option is C.

Given:

• $y_1(x, t) = 4\sin(kx - \omega t)$
• $y_2(x, t) = 2\sin\left(kx - \omega t + \frac{2\pi}{3}\right)$

Find: The amplitude and phase of the resultant wave.

Take $\theta = kx - \omega t$. Then

$$y = 4\sin\theta + 2\sin\left(\theta + \frac{2\pi}{3}\right)$$

Expand the second term:

$$\sin\left(\theta + \frac{2\pi}{3}\right) = \sin\theta \cos\frac{2\pi}{3} + \cos\theta \sin\frac{2\pi}{3}$$

So,

$$y = 4\sin\theta + 2\left(\sin\theta \cos\frac{2\pi}{3} + \cos\theta \sin\frac{2\pi}{3}\right)$$

Using

$$\cos\frac{2\pi}{3} = -\frac{1}{2}, \qquad \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$$

we get

$$y = 4\sin\theta + 2\left(-\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\right)$$ $$y = 4\sin\theta - \sin\theta + \sqrt{3}\cos\theta$$ $$y = 3\sin\theta + \sqrt{3}\cos\theta$$

Compare with

$$y = A\sin(\theta + \phi) = A\sin\theta \cos\phi + A\cos\theta \sin\phi$$

Thus,

$$A\cos\phi = 3, \qquad A\sin\phi = \sqrt{3}$$

Squaring and adding,

$$A^2 = 3^2 + (\sqrt{3})^2 = 9 + 3 = 12$$ $$A = 2\sqrt{3}$$

Now,

$$\tan\phi = \frac{A\sin\phi}{A\cos\phi} = \frac{\sqrt{3}}{3}$$

Hence,

$$\phi = \frac{\pi}{6}$$

Therefore, the resultant wave has amplitude $2\sqrt{3}$ and phase $\frac{\pi}{6}$. The correct option is C.