If x = A cos 4t + B sin 4t,then frac{d^2x}{dt | vedclass

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Question

(A) Given the equation: $x = A \cos 4t + B \sin 4t$ Differentiating with respect to $t$: $\frac{dx}{dt} = -4A \sin 4t + 4B \cos 4t$ Differentiating ag

Vedclass · Accepted Answer

$-16x$

Vedclass · Answer

(A) Given the equation: $x = A \cos 4t + B \sin 4t$ Differentiating with respect to $t$: $\frac{dx}{dt} = -4A \sin 4t + 4B \cos 4t$ Differentiating again with respect to $t$: $\frac{d^2x}{dt^2} = -16A \cos 4t - 16B \sin 4t$ Factoring out $-16$: $\frac{d^2x}{dt^2} = -16(A \cos 4t + B \sin 4t)$ Since $x = A \cos 4t + B \sin 4t$,we substitute $x$ into the equation: $\frac{d^2x}{dt^2} = -16x$

If $x = A \cos 4t + B \sin 4t$,then $\frac{d^2x}{dt^2} = $

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