The differential equation of the simple harmo | vedclass

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Question

(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$ Differentiating with respect to $t$: $\frac{dx}{dt} = -A n \sin

Vedclass · Accepted Answer

$\frac{d^2 x}{d t^2}+n^2 x=0$

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(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$ Differentiating with respect to $t$: $\frac{dx}{dt} = -A n \sin (nt + \alpha)$ ... (ii) Differentiating again with respect to $t$: $\frac{d^2x}{dt^2} = -An \frac{d}{dt} \sin (nt + \alpha)$ $\frac{d^2x}{dt^2} = -An^2 \cos (nt + \alpha)$ Substituting $x = A \cos (nt + \alpha)$ into the equation: $\frac{d^2x}{dt^2} = -n^2 x$ Therefore, the differential equation is: $\frac{d^2x}{dt^2} + n^2 x = 0$

The differential equation of the simple harmonic motion given by $x=A \cos (n t+\alpha)$ is

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