The resultant amplitude due to superposition of two waves Y 1-5 sin - t - kx and Y 2-5 cos- t - kx -150-A- 5B- 5 -3- 5 -2-3D- 5 -2-3

The correct option is A 5Given: Y_1=5 sin (ω t kx) and Y_2= 5 cos (ω t kx 150^∘)=5sin(ω t kx 240^∘) nbsp; Y=Y_1+Y_2 where Y is the resultant wave The amplitu ...

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Standard XII

Physics

Resultant Amplitude

The resultant...

Question

The resultant amplitude due to superposition of two waves
$Y_{1} = 5 sin (ω t - k x)$ and $Y_{2} = - 5 cos (ω t - k x - 150^{\circ})$

5

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5 \sqrt{3}

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5 \sqrt{2 - \sqrt{3}}

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5 \sqrt{2 + \sqrt{3}}

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Solution

The correct option is A $5$
Given:
$Y_{1} = 5 sin (ω t - k x)$ and $Y_{2} = - 5 cos (ω t - k x - 150^{\circ}) = 5 sin (ω t - k x - 240^{\circ})$
$Y = Y_{1} + Y_{2}$
where $Y$ is the resultant wave
The amplitude of this resultant wave is given by the formula
$A = \sqrt{(A_{1})^{2} + (A_{2})^{2} + 2 A_{1} A_{2} cos ϕ}$
Here,
$A_{1} = 5, A_{2} = 5$ and $ϕ = 240^{\circ}$
On substituting the values on the given formula
$A = \sqrt{(5)^{2} + (5)^{2} + 2 \times 5 \times 5 cos (150^{\circ})}$
$= \sqrt{25 + 25 - 2 \times 25 \times \frac{1}{2}}$ $(∵ cos 240^{\circ} = - \frac{1}{2})$
Hence, $A = 5$

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The resultant amplitude due to superposition of two waves Y1=5sin(ωt−kx) and Y2=−5cos(ωt−kx−150∘)

The resultant amplitude due to superposition of two waves
$Y_{1} = 5 sin (ω t - k x)$ and $Y_{2} = - 5 cos (ω t - k x - 150^{\circ})$