The displacement of a particle varies with ti | vedclass

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Question

(D) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.To find the amplitude,we rewrite the equation in the form $x = A \sin(\om

Vedclass · Accepted Answer

The motion is $SHM$ with amplitude $\sqrt{a^2 + b^2}$.

Vedclass · Answer

(D) Given the displacement equation: $x = a \sin \omega t + b \cos \omega t$.To find the amplitude,we rewrite the equation in the form $x = A \sin(\omega t + \phi)$.Let $a = A \cos \phi$ and $b = A \sin \phi$.Then $x = A \cos \phi \sin \omega t + A \sin \phi \cos \omega t = A \sin(\omega t + \phi)$.Squaring and adding the expressions for $a$ and $b$:$a^2 + b^2 = A^2 \cos^2 \phi + A^2 \sin^2 \phi = A^2(\cos^2 \phi + \sin^2 \phi) = A^2$.Thus,the amplitude $A = \sqrt{a^2 + b^2}$.Since the equation can be expressed as a single sine function,the motion is $SHM$ 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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