
When a damped harmonic oscillator completes $100$ oscillations, its amplitude is reduced to $\dfrac{1}{3}$ of its initial value. What will be its amplitude when it completes $200$ oscillations?
(A) $\dfrac{1}{5}$
(B) $\dfrac{2}{3}$
(C) $\dfrac{1}{6}$
(D) $\dfrac{1}{9}$
Answer
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Hint A particle is said to execute simple harmonic motion if the restoring force or acceleration on it is directly proportional to the displacement and its direction is opposite to displacement. An oscillating body will come to rest after executing a few oscillations. This is because of the resistance of air. These types of oscillations are called damped oscillations.
Formula used
$a = {a_0}{e^{ - bt}}$
Where, ${a_0}$ stands for the initial amplitude of the oscillations, $b$ stands for the damping constant and $t$ stands for the time.
Step by step solution
In the case of damped oscillations, the amplitude decreases with time. The amplitude at any particular instant can be given by the expression
$a = {a_0}{e^{ - bt}}$
Let the time period of each oscillation be $T$
For $100$ oscillations, the amplitude will decrease by$\dfrac{1}{3}$ i.e.
For $t = 100T$, $a = \dfrac{{{a_0}}}{3}$
$\therefore a = \dfrac{{{a_0}}}{3} = {a_0}{e^{ - 100Tb}}$ $\left( {\because t = 100T} \right)$…………………equation
After the completion of $200$oscillations,
From , $\dfrac{1}{3} = {e^{ - 100Tb}}$
Then, squaring both sides,
${\left( {{e^{ - 100Tb}}} \right)^2} = {\left( {\dfrac{1}{3}} \right)^2}$
$\therefore {e^{ - 200Tb}} = \dfrac{1}{9} $
Therefore, The answer is Option (D): $\dfrac{1}{9}$
Note
The frequency of damped oscillations is slightly less than the natural frequency. Almost all oscillations are damped oscillations, in one way or another. Friction is the main reason for the damping of oscillation in most cases. When the damping constant is very small i.e. $b < \sqrt {4mk} $ the decay of amplitude will be exponential. Such a system is called an underdamped system. If the damping constant $b = \sqrt {4mk} $ then the system will asymptotically reach the equilibrium, such a system is called a critically damped system. Shock absorbers of vehicles are an example of a critically damped system. When the damping constant $b < \sqrt {4mk} $ the system will approach equilibrium after a long time. Such a system is called an overdamped system.
Formula used
$a = {a_0}{e^{ - bt}}$
Where, ${a_0}$ stands for the initial amplitude of the oscillations, $b$ stands for the damping constant and $t$ stands for the time.
Step by step solution
In the case of damped oscillations, the amplitude decreases with time. The amplitude at any particular instant can be given by the expression
$a = {a_0}{e^{ - bt}}$
Let the time period of each oscillation be $T$
For $100$ oscillations, the amplitude will decrease by$\dfrac{1}{3}$ i.e.
For $t = 100T$, $a = \dfrac{{{a_0}}}{3}$
$\therefore a = \dfrac{{{a_0}}}{3} = {a_0}{e^{ - 100Tb}}$ $\left( {\because t = 100T} \right)$…………………equation
After the completion of $200$oscillations,
From , $\dfrac{1}{3} = {e^{ - 100Tb}}$
Then, squaring both sides,
${\left( {{e^{ - 100Tb}}} \right)^2} = {\left( {\dfrac{1}{3}} \right)^2}$
$\therefore {e^{ - 200Tb}} = \dfrac{1}{9} $
Therefore, The answer is Option (D): $\dfrac{1}{9}$
Note
The frequency of damped oscillations is slightly less than the natural frequency. Almost all oscillations are damped oscillations, in one way or another. Friction is the main reason for the damping of oscillation in most cases. When the damping constant is very small i.e. $b < \sqrt {4mk} $ the decay of amplitude will be exponential. Such a system is called an underdamped system. If the damping constant $b = \sqrt {4mk} $ then the system will asymptotically reach the equilibrium, such a system is called a critically damped system. Shock absorbers of vehicles are an example of a critically damped system. When the damping constant $b < \sqrt {4mk} $ the system will approach equilibrium after a long time. Such a system is called an overdamped system.
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