Question

17.Three waves of equal frequency having amplitude 10mm,4mm and 7mm arrive at a given point with successive phase difference is pie/2.The amplitude of the resulting wave in mm is given by----------

Answer 1

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Answer 2

Answer:

The amplitude of the resulting wave is 5 mm.

Explanation:

Given that,

Amplitude $a_{1} = 10 mm$

Amplitude $a_{2}=4 mm$

Amplitude $a_{3}=7 mm$

Phase difference $\phi=\dfrac{\pi}{2}$

Let the waves be

$x_{1}=10\ sin\ (\omega t)$

$x_{2}=4\ sin\ (\omega t+\dfrac{\pi}{2})$

$x_{2}=4\ cos\ (\omega t)$

$x_{3}=7\sin\ (\omega t+\pi)$

$x_{3}=-7\ sin\ (\omega t)$

Using superposition theorem

The resultant disturbance is given by

$x= x_{1}+x_{2}+x_{3}$

$x=10\ sin\ (\omega t)+4\ cos\ (\omega t)+(-7\ sin\ (\omega t))$

$x = 3\ sin\ (\omega t)+4\ cos\ (\omega t)$

Let R be the resultant amplitude

$R\ cos\phi= 3$

$R\ sin\ \phi=4$

$x= R\ cos\ \phi\ sin\ (\omega t)+R\ sin\ \phi\ cos\ (\omega t)$

$x= R\ sin\ (\omega t+\phi)$

$R= \sqrt{(3)^2+(4)^2}$

$R= 5\ mm$

Hence, The amplitude of the resulting wave is 5 mm.