When two progressive waves y1 = 4 sin (2x – 6t) and y2 = 3 sin (2x−6t−π2) are superimposed, the amplitude of the resultant wave is

When two progressive waves y1 = 4 sin (2x – 6t) and y2 = 3 sin \left(\right. 2 x - 6 t - \frac{\pi}{2} \left.\right) are superimposed, the amplitude of the resultant wave is

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Prograssive Wave Equation

Reflection Refraction and Superposition of Wave

Question

When two progressive waves y₁ = 4 sin (2x – 6t) and

y₂ = 3 sin $(2 x - 6 t - \frac{π}{2})$

are superimposed, the amplitude of the resultant wave is

$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cosθ}$

A₁ = 4

A₂ = 3 $\Rightarrow$ A = 5

$ϕ$ = – $π$ /2

When two waves superimpose, their displacements add up to give the resultant wave. The amplitude of the resultant wave depends on the phase difference between the two waves.

Given waves:

y₁ = 4 sin(2x - 6t)

y₂ = 3 sin $(2 x - 6 t - \frac{π}{2})$

Step 1: Identify the amplitudes and phase difference

Amplitude of first wave, A₁ = 4

Amplitude of second wave, A₂ = 3

The phase difference (φ) between the two waves is the difference in their phase angles. The second wave has an additional phase of -π/2 compared to the first wave.

Therefore, phase difference φ = π/2 radians (the magnitude of the difference).

Step 2: Use the formula for the amplitude of the resultant wave

The amplitude A of the resultant wave when two waves of amplitudes A₁ and A₂ with a phase difference φ superimpose is given by:

$A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos φ}$

Step 3: Substitute the values

A₁ = 4, A₂ = 3, φ = π/2

cos(π/2) = 0

Therefore, the equation simplifies to:

$A = \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times 0} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$

Final Answer: The amplitude of the resultant wave is 5 units.

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

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