
In damped oscillations amplitude after 50 oscillations is 0.8 a$_0$ where a$_0$ is initial amplitude. Then amplitude after 150 oscillations will be
A. $\dfrac{{0.8}}{3}{a_0}$
B. 0.24${a_0}$
C. 0.512${a_0}$
D. 0.8${a_0}$
Answer
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Hint: The general equation for the amplitude(x) in damped oscillation is given as $x = {a_0}\exp \left( { - \alpha t} \right)$, where a$_0$ is the initial amplitude, $\alpha $ is a constant and t is time. One oscillation is completed in one time period(T). Therefore on substituting the time taken in the two cases (50T and 150T) and comparing the equations we get the desired amplitude(x).
Complete step by step answer:
Given the initial amplitude is a$_0$.
The general equation for amplitude(x) of damped oscillation is given as,
$x = {a_0}\exp \left( { - \alpha t} \right)$ -----(1)
where ${a_0}$ is the initial amplitude, $\alpha $ is a constant and t is time.
The time taken to complete one oscillation is one time period(T).
Case 1: Given the time taken is equal to 50 oscillation$ = 50T$.Amplitude is equal to 0.8 ${a_0}$.
Substituting the given values in equation (1) we get,
$0.8{a_0} = {a_0}\exp ( - \alpha.50T)$
$ \Rightarrow \exp ( - \alpha.50T) = 0.8$ ------------(2)
Case 2: Given the time taken is equal to 50 oscillation$ = 150T$.
Let the unknown amplitude be equal to x.
Substituting the given values in equation (1) we get,
$
x = {a_0}\exp ( - \alpha.150T) \\
\Rightarrow x = {a_0}{\left[ {\exp \left( { - \alpha.50T} \right)} \right]^3} \\
$
Substituting value from equation (2), $\exp ( - \alpha.50T) = 0.8$ we get
$\therefore x = {a_0}{\left( {0.8} \right)^3}$
Therefore the required amplitude is $x = {a_0}{\left( {0.8} \right)^3}$$ = 0.512{a_0}$.
Note:One should know the general formula for amplitude of a particle in damped oscillation motion to solve such a type of question, $x = {a_0}\exp \left( { - \alpha t} \right)$, where a$_0$ is the initial amplitude, $\alpha $ is a constant and t is time.A simple harmonic motion for oscillating motion in which a damping force reduces the amplitude of the motion is known as damped oscillatory motion.
Complete step by step answer:
Given the initial amplitude is a$_0$.
The general equation for amplitude(x) of damped oscillation is given as,
$x = {a_0}\exp \left( { - \alpha t} \right)$ -----(1)
where ${a_0}$ is the initial amplitude, $\alpha $ is a constant and t is time.
The time taken to complete one oscillation is one time period(T).
Case 1: Given the time taken is equal to 50 oscillation$ = 50T$.Amplitude is equal to 0.8 ${a_0}$.
Substituting the given values in equation (1) we get,
$0.8{a_0} = {a_0}\exp ( - \alpha.50T)$
$ \Rightarrow \exp ( - \alpha.50T) = 0.8$ ------------(2)
Case 2: Given the time taken is equal to 50 oscillation$ = 150T$.
Let the unknown amplitude be equal to x.
Substituting the given values in equation (1) we get,
$
x = {a_0}\exp ( - \alpha.150T) \\
\Rightarrow x = {a_0}{\left[ {\exp \left( { - \alpha.50T} \right)} \right]^3} \\
$
Substituting value from equation (2), $\exp ( - \alpha.50T) = 0.8$ we get
$\therefore x = {a_0}{\left( {0.8} \right)^3}$
Therefore the required amplitude is $x = {a_0}{\left( {0.8} \right)^3}$$ = 0.512{a_0}$.
Note:One should know the general formula for amplitude of a particle in damped oscillation motion to solve such a type of question, $x = {a_0}\exp \left( { - \alpha t} \right)$, where a$_0$ is the initial amplitude, $\alpha $ is a constant and t is time.A simple harmonic motion for oscillating motion in which a damping force reduces the amplitude of the motion is known as damped oscillatory motion.
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