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The amplitudes of a damped harmonic oscillator after 2 and 4 seconds are $A_1$ and $A_2$ respectively. If…
The amplitudes of a damped harmonic oscillator after 2 and 4 seconds are $A_1$ and $A_2$ respectively. If the initial amplitude of the oscillator is $\mathrm{A}_0$, then
$\mathrm{A}_1=\sqrt{\mathrm{A}_0 \mathrm{~A}_2}$ $\mathrm{A}_2=\sqrt{\mathrm{A}_0 \mathrm{~A}_1}$ $\mathrm{A}_0=\sqrt{\mathrm{A}_1 \mathrm{~A}_2}$ $\mathrm{A}_1=\frac{\mathrm{A}_0+\mathrm{A}_2}{2}$ Solution
In damping harmonic oscillator, amplitude
$
\begin{aligned}
& A=A_0 e^{-b t} \\
& \therefore A_1=A_0 e^{-b \times 2} \Rightarrow \frac{A_1}{A_0}=e^{-2 b}
\end{aligned}
$
And
$
\begin{aligned}
& A_2=A_0 e^{-b \times 4} \Rightarrow \frac{A_2}{A_0}=e^{-4 b}=\left(e^{-2 b}\right)^2=\left(\frac{A_1}{A_0}\right)^2 \\
& \text { or, } \frac{A_2}{A_0}=\left(\frac{A_1}{A_0}\right)^2=\frac{A_1^2}{A_0^2} \text { or, } \frac{A_2}{A_1^2}=\frac{1}{A_0} \therefore A_1=\sqrt{A_0 A_2}
\end{aligned}
$
Asked in: AP EAMCET 2023 (15 May Shift 1)
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