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:

  1. 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, $\ln (2)=0.693$

$$ \begin{aligned} \therefore & t & =10 \times 0.693=6.93 s \\ \text { or } & t & \approx 7 s \end{aligned} $$



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