Question

In damped oscillations, damping force is directly proportional to speed to oscillator. If amplitude becomes half of its maximum value in 1s, then after 2s amplitude will be (initial amplitude = $A_0$)

A. ${\displaystyle \frac{1}{4}}{A}_0$
B. ${\displaystyle \frac{1}{2}}{A}_0$
C. ${\displaystyle \frac{1}{5}}{A}_0$
.D. ${\displaystyle \frac{1}{7}}{A}_0$

Answer 1

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 try to find the time t and finally using that time value find the value of amplitude after 2 sec time.



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


Complete answer:
For case 1: t = 1 sec
Given the amplitude at time t = 1 sec becomes half of its initial amplitude.
$A={\displaystyle \frac{A_0}{2}}$
Putting this value in the formula, we get;
${\displaystyle \frac{A_0}{2}}=A_0{e}^{-\alpha }$
After solving, we get: $e^{-\alpha }={\displaystyle \frac{1}{2}}$ (equation 1)
For case 2: t=2 sec
$A=A_0{e}^{-2\alpha }$
Putting the value from equation 1, we get;
$A=A_0{\left({\displaystyle \frac{1}{2}}\right)}^2$
After solving, we get;
$A={\displaystyle \frac{A_0}{4}}$
Therefore, amplitude at time t = 2 sec will become 1/4 times of the initial amplitude.

Hence, the correct answer is Option(A).


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 sec and t = 2 sec. Also use the value of the $e^{-\alpha }={\displaystyle \frac{1}{2}}$to get the required result.