Simple Harmonic Motion 3 Question 1
1. The displacement of a damped harmonic oscillator is given by $x(t)=e^{-0.1 t} \cos (10 \pi t+\varphi)$.
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: 1. (d)
Solution:
- Given, displacement is
$$ x(t)=e^{-0.1 t} \cos (10 \pi t+\varphi) $$
Here, amplitude of the oscillator is
$$ A=e^{-0.1 t} $$
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)]
At
So, Eq. (i) can be written as
$$ \begin{aligned} e^{-0.1 t} & =\frac{1}{2} \\ \text { or } \quad e^{0.1 t} & =2 \end{aligned} $$
$$ \begin{aligned} & \text { or } \quad 0.1 t=\ln (2) \\ & \text { or } \\ & t=\frac{1}{0.1} \ln (2)=10 \ln (2) \end{aligned} $$
Now, $\ln (2)=0.693$
$$ \begin{aligned} \therefore & t & =10 \times 0.693=6.93 s \\ \text { or } & t & \approx 7 s \end{aligned} $$
