AP EAMCET · PHYSICS · Oscillations
The time taken for the amplitude of vibrations of a damped oscillator to drop to \(25 \%\) of its initial value is \(t\). The time taken for its mechanical energy to drop to \(50 \%\) of its initial mechanical energy is
- A \(\mathrm{t}\)
- B \(\frac{\mathrm{t}}{2}\)
- C \(\frac{t}{4}\)
- D \(\frac{t}{8}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{t}}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & A=A_0 e^{-\frac{b t}{2 m}} \\ & A=25 \% \text { of } A_0=\frac{A_0}{4} \\ & \frac{A_0}{4}=A_0 e^{-\frac{b t}{2 m}} \\ & (2)^{-2}=e^{-\frac{b t}{2 m}} \\ & \frac{b t}{2 m}=2 \ln 2 \Rightarrow t=\frac{42 m \ell n 2}{b} ... (1) \\ & E(t)=E(0) e^{-\frac{b t}{2 m}}…
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