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If the amplitude of a lightly damped oscillator decreases by $1.5 \%$ then the mechanical energy of the oscillator lost in each cycle is
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$3 \%$
We know that, mechanical energy of oscillator lost in each cycle is proportional to the square of amplitude $(A)$. Then fractional change in energy
$\frac{\Delta E}{E} \simeq \frac{d E}{E}=\frac{d A^2}{A^2}$
$\begin{aligned} & =\frac{2 A d A}{A^2}=2 \frac{d A}{A} \\ & =2 \times 1.5 \%=3 \%\end{aligned}$
$\frac{\Delta E}{E} \simeq \frac{d E}{E}=\frac{d A^2}{A^2}$
$\begin{aligned} & =\frac{2 A d A}{A^2}=2 \frac{d A}{A} \\ & =2 \times 1.5 \%=3 \%\end{aligned}$
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