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If $x = A \cos 4t + B \sin 4t$, then $\frac{d^2x}{dt^2}$ is equal to: |
$x$ $-x$ $16x$ $-16x$ |
$-16x$ |
The correct answer is Option (4) → $-16x$ ## $x = A \cos 4t + B \sin 4t$ $\frac{dx}{dt} = -4A \sin 4t + 4B \cos 4t$ $\frac{d^2x}{dt^2} = -16A \cos 4t - 16B \sin 4t$ $= -16(A \cos 4t + B \sin 4t)$ $∴\frac{d^2x}{dt^2} = -16x$ |
