The displacement of a particle varies with ti | vedclass

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Question

(B) The given equation is $x = a \sin \omega t + b \cos \omega t$.To find the amplitude, we can rewrite this in the form $x = A \sin(\omega t + \phi)$

Vedclass · Accepted Answer

The motion is simple harmonic motion with an amplitude $\sqrt{a^2 + b^2}$

Vedclass · Answer

(B) The given equation is $x = a \sin \omega t + b \cos \omega t$.To find the amplitude, we can rewrite this in the form $x = A \sin(\omega t + \phi)$.Let $a = A \cos \phi$ and $b = A \sin \phi$.Squaring and adding these equations: $a^2 + b^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.Thus, the amplitude $A = \sqrt{a^2 + b^2}$.The equation becomes $x = \sqrt{a^2 + b^2} \sin(\omega t + \phi)$, which represents simple harmonic motion $(SHM)$ with amplitude $\sqrt{a^2 + b^2}$.

The displacement of a particle varies with time according to the relation $x = a \sin \omega t + b \cos \omega t$.

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