## Question

When two progressive waves **y _{1}=4 sin(2x–6t)** and

**y**are superimposed, the amplitude of the resultant wave is

_{2 }= 3sin(2x−6t−(π/2))## 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 **y _{1} = 4sin(2x–6t)**, and

**y**with the formula

_{2 }= 3sin(2x−6t−(π/2))**y = A sin (ωt + ϕ)**. We will have

**y _{1} = 4sin(2x–6t)** which implies that

**A**

_{1}= 4**y _{2 }= 3sin(2x−6t−(π/2))** which also shows that

**A**

_{2}= 3It’s also obvious that **ϕ =** **π/2**

But **π** = 180^{0}

Therefore, **π/2 = 180 ^{0} / 2 =90^{0}**

**Unrevealed value:**

Resultant Amplitude, A_{R} = ?

**Formula: **

We will use A_{R} = √(A_{1}^{2} + A_{2}^{2}) + 2A_{1}A_{2} cosϕ

#### Solution

We will now insert our data into A_{R} = √(A_{1}^{2} + A_{2}^{2}) + 2A_{1}A_{2} cosϕ to solve the problem

A_{R} = √(A_{1}^{2} + A_{2}^{2}) + 2A_{1}A_{2} cosϕ = A_{R} = √(4^{2} + 3^{2}) + 2 x 4 x 3 x cos90^{0}

But cos90^{0} = 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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