## 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

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, AR = ?

Formula:

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

#### Solution

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

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

But cos900 = 0

Which implies that

AR = √(16 + 9) + 2 x 4 x 3 x 0

Thus

AR = √(25) + 0

Therefore

AR = √(25) + 0 = √(25) = 5 m

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

y = A sin (ωt + ϕ)

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