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

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,$$R = \sqrt{A^2 + B^2 + 2 AB cos 90^{\circ}}$$$$= \sqrt{A^2 + B^2}$$Physics

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