When two progressive waves and are superimposed, the amplitude of the resultant wave is [2010]

Question

When two progressive waves $y_{1} = 4 \sin (2 x - 6 t)$ and $y_{2} = 3 \sin (2 x - 6 t - \frac{π}{2})$ are superimposed, the amplitude of the resultant wave is [2010]

Smart Achievers · Accepted Answer

(5)

Here the phase difference between the two waves, $ϕ = \frac{π}{2}$

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

$= \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{2}}$

$= \sqrt{16 + 9 + 0} = 5$

Solution

Q.
When two progressive waves $y_{1} = 4 \sin (2 x - 6 t)$ and $y_{2} = 3 \sin (2 x - 6 t - \frac{π}{2})$ are superimposed, the amplitude of the resultant wave is [2010]

Ans.
(5)

Here the phase difference between the two waves, $ϕ = \frac{π}{2}$

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

$= \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{2}}$

$= \sqrt{16 + 9 + 0} = 5$

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Solution

Q. When two progressive waves y1=4sin(2x-6t) and y2=3sin(2x-6t-π2) are superimposed, the amplitude of the resultant wave is [2010]

Ans. (5) Here the phase difference between the two waves, ϕ=π2 Resultant amplitude, A=A12+A22+2A1A2cosϕ =42+32+2×4×3×cosπ2 =16+9+0=5

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Q.
When two progressive waves $y_{1} = 4 \sin (2 x - 6 t)$ and $y_{2} = 3 \sin (2 x - 6 t - \frac{π}{2})$ are superimposed, the amplitude of the resultant wave is [2010]

Ans.
(5)

Here the phase difference between the two waves, $ϕ = \frac{π}{2}$

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

$= \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3 \times \cos \frac{π}{2}}$

$= \sqrt{16 + 9 + 0} = 5$