The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\dfrac{1}{X}$ times the original, where X is

A. $2 \times 3$
B. ${2^3}$
C. ${3^2}$
D. $3 \times {2^2}$

Hint:In case of damped oscillations, first try to find the relation between the amplitude at time t and the initial amplitude. Then put the values of both the amplitude and the time t is given and finally using that time value find the value of amplitude after 3 minutes time.



Formula used
$A = {A_0}{e^{ - \alpha t}}$
Where, A is the amplitude at a given time t.
And ${A_0}$is the initial amplitude.


Complete answer:

For case 1: t = 1 minute
Given the amplitude at time t = 1 minute becomes half of its initial amplitude.
$A = \dfrac{{{A_0}}}{2}$
Putting this value in the formula, we get;
$\dfrac{{{A_0}}}{2} = {A_0}{e^{ - \alpha }}$
After solving, we get: ${e^{ - \alpha }} = \dfrac{1}{2}$ (equation 1)
For case 2: t=3 minutes
$\dfrac{{{A_0}}}{X} = {A_0}{e^{ - 3\alpha }}$
${e^{ - 3\alpha }} = \dfrac{1}{X}$ (equation 2)
Solving equation 1 and 2, we get;
$\dfrac{1}{X} = {({e^{ - \alpha }})^3}$
$\dfrac{1}{X} = {\left( {\dfrac{1}{2}} \right)^3} = \dfrac{1}{8}$
By solving, we get;
$X = 8 = {2^3}$

Hence, the correct answer is Option(B).



Note:Be careful about the change in amplitude according to the time given and how many times it becomes of the initial amplitude or the maximum amplitude. Use the same formula for both the case at t = 1 minutes and t = 3 minutes. Finally solving both cases together gets the value of X.