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Question

The displacement of a particle executing simple harmonic motion is given by $$y = A_0 + A \sin \omega t + B \cos \omega t$$
Then the amplitude of its oscillation is given by :


A
A0+A2+B2
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B
A2+B2
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C
A20+(A+B)2
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D
A+B
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Solution

The correct option is C $$\sqrt{A^2 + B^2}$$
Given,
$$y = A_0 + A sin \omega t + B sin \omega t$$
Equation of  SHM
$$y' = y - A_0 = A sin \omega  t + B cos \omega t$$
Resultant amplitude,
$$R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}$$
$$= \sqrt{A^2 + B^2}$$

1259345_1618724_ans_a949543ff460417897ace0f485dd09b6.png

Physics

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