Three waves of amplitudes 7 mm , 3 mm and 11 mm with a successive phase difference of π 2 are super -imposed. The amplitude of the resulting wave will be

Let us say that the first wave has a zero phase, then the second one is ahead of the first by a phase of π 2 and so the third one is ahead of the first one by a phase of π . So the first and the third are opposite to each other. Hence, A r = ( 11 − 7 ) 2 + ( 3 ) 2 A r = 5 mm

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PhysicsSound WaveInterferenceMedium2 minPYQ_2020

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Three waves of amplitudes $7 mm$ , $3 mm$ and $11 mm$ with a successive phase difference of $\frac{π}{2}$ are super -imposed. The amplitude of the resulting wave will be

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Solution:

Let us say that the first wave has a zero phase, then the second one is ahead of the first by a phase of $\frac{π}{2}$ and so the third one is ahead of the first one by a phase of $π$ . So the first and the third are opposite to each other. Hence,

$A_{r} = \sqrt{{(11 - 7)}^{2} + {(3)}^{2}}$

$A_{r} = 5 mm$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Interference

⏱ 2mℹ️ Source: PYQ_2020

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