A particle executes SHM given by the equation $y = 3 \sin(\omega t) + 4 \cos(\omega t)$ (in cm). Wha

Question

A particle executes SHM given by the equation $y = 3 \sin(\omega t) + 4 \cos(\omega t)$ (in cm). What is the amplitude of this oscillation?

Professor PC Thomas Classes · Answer

The two SHMs are perpendicular in phase (sine and cosine). The resultant amplitude is $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ cm.