Simple Harmonic Motion Ques 17
- The displacement of a damped harmonic oscillator is given by $x(t)=e^{-0.1 t} \cos (10 \pi t+\phi)$.
Here, $t$ is in seconds.
The time taken for its amplitude of vibration to drop to half of its initial value is close to
(2019 Main, 10 April I)
(a) $27$ $s$
(b) $13$ $s$
(c) $4$ $s$
(d) $7$ $s$
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Answer:
Correct Answer: 17.(d)
Solution:
Formula:
- Given, displacement is
$ x(t)=e^{-0.1 t} \cos (10 \pi t+\phi) $
Here, amplitude of the oscillator is
$ A=e^{-0.1 t} $ $\quad$ …….(i)
Let it takes $t$ seconds for amplitude to be dropped by half.
At
$ \begin{aligned} & t=0 \Rightarrow A=1 \\ & t=t \Rightarrow A^{\prime}=\frac{A}{2}=\frac{1}{2} \end{aligned} $
[from Eq. (i)]
So, Eq. (i) can be written as
$ \begin{aligned} e^{-0.1 t} & =\frac{1}{2} \\ \text { or } \quad e^{0.1 t} & =2 \end{aligned} $
$ \begin{aligned} & \text { or } \quad 0.1 t=\ln (2) \\ & \text { or } \quad t=\frac{1}{0.1} \ln (2)=10 \ln (2) \end{aligned} $
Now, $\ln (2)=0.693$
$ \begin{aligned} \therefore & \quad t =10 \times 0.693=6.93 s \\ \text { or } & \quad t \approx 7 s \end{aligned} $
BETA
