When two displacements represented by y_1 = a | vedclass

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Question

(C) Given the two displacements:$y_1 = a \sin(\omega t)$$y_2 = b \cos(\omega t) = b \sin(\omega t + \frac{\pi}{2})$The resultant displacement $y = y_1

Vedclass · Accepted Answer

simple harmonic with amplitude $\sqrt{a^2 + b^2}$

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(C) Given the two displacements:$y_1 = a \sin(\omega t)$$y_2 = b \cos(\omega t) = b \sin(\omega t + \frac{\pi}{2})$The resultant displacement $y = y_1 + y_2$ is given by:$y = a \sin(\omega t) + b \cos(\omega t)$To find the amplitude,we can write this in the form $y = A \sin(\omega t + \phi)$,where the resultant amplitude $A$ is:$A = \sqrt{a^2 + b^2 + 2ab \cos(\frac{\pi}{2})}$Since $\cos(\frac{\pi}{2}) = 0$,the amplitude is:$A = \sqrt{a^2 + b^2}$Thus,the resultant motion is simple harmonic with an amplitude of $\sqrt{a^2 + b^2}$.

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

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