Two waves are simultaneously passing through a string and their equations are : y1 = A1 sin k(x - vt), y2 = A2 sin k(x - vt + x0). Given amplitudes A1 = 12 mm and A2 = 5 mm, x0 = 3.5 cm and wave number k = 6.28 cm - 1. The amplitude of resulting wave will be ................ mm.

Waves PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q63 with Solution

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Correct answer: 7

Given: $y_1 = A_1 \sin k(x-vt), \qquad y_2 = A_2 \sin k(x-vt+x_0)$

where $A_1 = 12\text{ mm}, \quad A_2 = 5\text{ mm}, \quad x_0 = 3.5\text{ cm}, \quad k = 6.28\text{ cm}^{-1}$

Let $\theta = k(x-vt)$

Then the waves become $y_1 = A_1 \sin\theta, \qquad y_2 = A_2 \sin(\theta + kx_0)$

So the phase difference is $\phi = kx_0$

$\phi = kx_0 = 6.28 \times 3.5 = 21.98 \text{ rad}$

Now, $21.98 \approx 7\pi$

since $7\pi \approx 21.99$

Therefore, $\phi \approx 7\pi$

and $\cos\phi = \cos(7\pi) = -1$

Thus the two waves are in opposite phase.

For two waves of same frequency and wave number, $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$

Substitute values: $A = \sqrt{12^2 + 5^2 + 2(12)(5)(-1)}$

$A = \sqrt{144 + 25 - 120}$

$A = \sqrt{49} = 7\text{ mm}$

The amplitude of the resulting wave is $\boxed{7\text{ mm}}$

Stored correct answer = 7

Our derived answer = 7

So the answer matches the stored correct answer.