Simple Harmonic Motion Ques 60
- A damped harmonic oscillator has a frequency of $5$ oscillations per second. The amplitude drops to half its value for every $10$ oscillations. The time it will take to drop to $\frac{1}{1000}$ of the original amplitude is close to
(2019 Main, 8 April II)
(a) $20$ $ s$
(b) $50$ $ s$
(c) $100$ $ s$
(d) $10$ $ s$
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Answer:
Correct Answer: 60.(a)
Solution:
Formula:
- Given, frequency of oscillations is $f=5$ $ osc$ $ s^{-1}$
$\Rightarrow$ Time period of oscillations is $T=\frac{1}{f}=\frac{1}{5} s$
So, time for 10 oscillations is $=\frac{10}{5}=2 s$
Now, if $A _0=$ initial amplitude at $t=0$ and $\gamma=$ damping factor, then for damped oscillations, amplitude after $t$ second is given as
$ A=A _0 e^{-\gamma t} $
$\therefore$ After $2 s$,
$ \begin{aligned} & \frac{A _0}{2}=A _0 e^{-\gamma(2)} \Rightarrow 2=e^{2 \gamma} \\ \Rightarrow \quad & \gamma=\frac{\log 2}{2} \quad …….(i) \end{aligned} $
Now, when amplitude is $\frac{1}{1000}$ of initial amplitude, i.e.
$\frac{A _0}{1000}=A _0 e^{-\gamma t} $
$\Rightarrow \quad \log (1000)=\gamma t $
$\Rightarrow \quad \log \left(10^{3}\right)=\gamma t $
$3 \log 10=\gamma t $
$\Rightarrow \quad t=\frac{2 \times 3 \log 10}{\log 2} \quad $ [using Eq. (i)]
$\Rightarrow \quad t=19.93 s $
$\text { or } \quad t \approx 20 s$
BETA
