Three harmonic waves- having equal frequency f and intensity I0- have phase angles 0-4 and -4 respectively- When they are superimposed- the intensity of the resultant wave is close to-

Let the amplitude of each wave is a_1=a_2=a_3=a_0. Now, ϕ_1=0, ϕ_2=π4, ϕ_3= π4 The resultant amplitude due to superposition of second and third wave is, a_r= ...

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Standard XII

Physics

Principle of Superposition

Three harmoni...

Question

Three harmonic waves, having equal frequency $f$ and intensity $I_{0}$ , have phase angles $0, \frac{π}{4} and - \frac{π}{4}$ respectively. When they are superimposed, the intensity of the resultant wave is close to:

5.8 I_{0}

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0.2 I_{0}

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3 I_{0}

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I_{0}

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Solution

The correct option is A $5.8 I_{0}$
Let the amplitude of each wave is $a_{1} = a_{2} = a_{3} = a_{0}$ .

Now, $ϕ_{1} = 0, ϕ_{2} = \frac{π}{4}, ϕ_{3} = - \frac{π}{4}$

The resultant amplitude due to superposition of second and third wave is,

$a_{r} = \sqrt{a_{2}^{2} + a_{3}^{2} + 2 a_{2} a_{3} cos (π / 2)}$

$= \sqrt{a_{0}^{2} + a_{0}^{2} + 2 a_{0}^{2} (0)}$

$∴ a_{r} = a_{0} \sqrt{2}$

Also, $tan δ = \frac{a_{2} sin ϕ_{2} + a_{2} sin ϕ_{3}}{a_{2} cos ϕ_{2} + a_{3} cos ϕ_{3}}$

$= \frac{(\frac{a_{0}}{\sqrt{2}} - \frac{a_{0}}{\sqrt{2}})}{(\frac{a_{0}}{\sqrt{2}} + \frac{a_{0}}{\sqrt{2}})} = 0$

The resultant amplitude of the above wave $(a_{r} = a_{0} \sqrt{2}, δ = 0)$ and the first wave $(a_{1} = a_{0}, ϕ_{1} = 0)$ is,

$A = \sqrt{a_{1}^{2} + a_{r}^{2} + 2 a_{1} a_{r} cos (0)}$

$= \sqrt{a_{0}^{2} + 2 a_{0}^{2} + 2 a_{0} (a_{0} \sqrt{2})} = a_{0} \sqrt{5.828}$

Now, the intensity is directly proportional to the square of the amplitude,

i.e. $I \propto a^{2}$

$\Rightarrow \frac{I_{0}}{I_{R}} = \frac{a_{0}^{2}}{(a_{0} \sqrt{5.828})^{2}}$

$∴ I_{R} = 5.828 I_{0} \approx 5.8 I_{0}$

Alternate solution:

Consider the phasor diagram of the given point, at which, the three waves are arriving simultaneously.

By the geometry of the figure, $a_{r} = a_{0} + a_{0} \sqrt{2}$

As, the intensity is directly proportional to the square of the amplitude,

$\Rightarrow \frac{I_{0}}{I_{R}} = \frac{a_{0}^{2}}{(a_{0} \sqrt{5.828})^{2}}$

$∴ I_{R} = 5.828 I_{0} \approx 5.8 I_{0}$

Hence, $(A)$ is the correct answer.

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Three harmonic waves, having equal frequency f and intensity I0, have phase angles 0,π4 and −π4 respectively. When they are superimposed, the intensity of the resultant wave is close to:

Three harmonic waves, having equal frequency $f$ and intensity $I_{0}$ , have phase angles $0, \frac{π}{4} and - \frac{π}{4}$ respectively. When they are superimposed, the intensity of the resultant wave is close to: