The displacement of a particle executing simple harmonic motion is given by y=A0+Asinωt+Bcosωty = A_0 + A\sin\omega t + B\cos\omega ty=A0+Asinωt+Bcosωt. Then the amplitude of its oscillation is given by :
A0+A2+B2A_0 + \sqrt{A^2 + B^2}A0+A2+B2
A2+B2\sqrt{A^2 + B^2}A2+B2
A02+(A+B)2A_0^2 + (A+B)^2A02+(A+B)2
A+BA + BA+B
Given y=A0+Asinωt+Bcosωty = A_0 + A\sin\omega t + B\cos\omega ty=A0+Asinωt+Bcosωt. Equate SHM: y′=y−A0=Asinωt+Bcosωty' = y - A_0 = A\sin\omega t + B\cos\omega ty′=y−A0=Asinωt+Bcosωt. Resultant amplitude R=A2+B2+2ABcos90∘=A2+B2R = \sqrt{A^2 + B^2 + 2AB\cos 90^\circ} = \sqrt{A^2 + B^2}R=A2+B2+2ABcos90∘=A2+B2.
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