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Question

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $\frac{dy}{dt} = -A \omega \sin \omega t + B \ome

Vedclass · Accepted Answer

$\frac{d^2 y}{d t^2} + \omega^2 y = 0$

Vedclass · Answer

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $\frac{dy}{dt} = -A \omega \sin \omega t + B \omega \cos \omega t$ Differentiating again with respect to $t$: $\frac{d^2y}{dt^2} = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t$ Factoring out $-\omega^2$: $\frac{d^2y}{dt^2} = -\omega^2 (A \cos \omega t + B \sin \omega t)$ Since $y = A \cos \omega t + B \sin \omega t$,we substitute $y$ into the equation: $\frac{d^2y}{dt^2} = -\omega^2 y$ Rearranging the terms gives: $\frac{d^2y}{dt^2} + \omega^2 y = 0$

The differential equation obtained by eliminating $A$ and $B$ from $y = A \cos \omega t + B \sin \omega t$ is:

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