when two progressive waves y1=4 sin(2x-6t)

Question

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

Answer

The final answer to the above question is 5 meters (5m).

Explanation to the Above Answer

Data: Important information from the question

For us to be able to fully understand how to extract our data, we must remember the formula for amplitude is

y = A sin (ωt + ϕ)

Where y = displacement of the wave, A = amplitude, ω = angular velocity, t = time, and ϕ = phase angle

From the above question, when we compare y₁ = 4sin(2x–6t), and y₂ = 3sin(2x−6t−(π/2)) with the formula y = A sin (ωt + ϕ). We will have

y₁ = 4sin(2x–6t) which implies that A₁ = 4

y₂ = 3sin(2x−6t−(π/2)) which also shows that A₂ = 3

It’s also obvious that ϕ = π/2

But π = 180⁰

Therefore, π/2 = 180⁰ / 2 =90⁰

Unrevealed value:

Resultant Amplitude, A_R = ?

Formula:

We will use A_R = √(A₁² + A₂²) + 2A₁A₂ cosϕ

Solution

We will now insert our data into A_R = √(A₁² + A₂²) + 2A₁A₂ cosϕ to solve the problem

A_R = √(A₁² + A₂²) + 2A₁A₂ cosϕ = A_R = √(4² + 3²) + 2 x 4 x 3 x cos90⁰

But cos90⁰ = 0

Which implies that

A_R = √(16 + 9) + 2 x 4 x 3 x 0

Thus

A_R = √(25) + 0

Therefore

A_R = √(25) + 0 = √(25) = 5 m

Therefore, the amplitude of the resultant wave is 5 meters.

y = A sin (ωt + ϕ)

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