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(A) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$.Differentiating with respect to time $t$, we get $\frac{dx}{dt} = A

Vedclass · Accepted Answer

$\frac{dx}{dt} = i \omega \sqrt{x^2 - A^2}$

Vedclass · Answer

(A) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$.Differentiating with respect to time $t$, we get $\frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.Using the identity $\cos 	heta = \sqrt{1 - \sin^2 	heta}$, we have $\frac{dx}{dt} = A \omega \sqrt{1 - \sin^2(\omega t + \phi)}$.Substituting $\sin(\omega t + \phi) = \frac{x}{A}$, we get $\frac{dx}{dt} = A \omega \sqrt{1 - \frac{x^2}{A^2}}$.Simplifying this, $\frac{dx}{dt} = \omega \sqrt{A^2 - x^2}$.Since $A^2 - x^2 = -(x^2 - A^2)$, we can write $\frac{dx}{dt} = \omega \sqrt{-(x^2 - A^2)}$.Using $i = \sqrt{-1}$, we get $\frac{dx}{dt} = i \omega \sqrt{x^2 - A^2}$.

Which of the following equations represents a simple harmonic motion? $(\omega$ is angular frequency, $A$ is amplitude of oscillation and $i = \sqrt{-1})$

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