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# The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decreases to $\alpha$ times its original magnitude where $\alpha$ equalsA: 0.81B: 0.729C: 0.6D: 0.7

Last updated date: 20th Jun 2024
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Hint: Amplitude of a damped oscillator decreases exponentially. For this first calculate the amplitude of the damped oscillator when it decreases o.9 times at 5 sec. Similarly find the decrease in amplitude at 10s using the concept amplitude decreases exponentially. By comparing both the equations we will get the value of $\alpha$ .
Formula used:
$A={{A}_{0}}{{e}^{-kt}}$
where, A is the new amplitude.
${A_0}$ is the initial amplitude.
t is the time taken to decrease the amplitude.
k is the damping constant.

The amplitude of oscillations gradually decreases to zero as a result of frictional forces, arising due to the viscosity of the medium in which the oscillator is moving. The motion of the oscillator is damped by friction and therefore is called a damped harmonic oscillator. .
If damping is taken into account then a harmonic oscillator experiences a restoring force and a damping force proportional to the velocity.
Then the equation of motion of the damped harmonic oscillator is given by,
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-\gamma \dfrac{dx}{dt}-Cx$
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+\gamma \dfrac{dx}{dt}+Cx=0$
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+2k\dfrac{dx}{dt}+{{\omega }_{0}}x=0$
where, $2k=\dfrac{\gamma }{m}$ and ${{\omega }_{0}}=\sqrt{\dfrac{C}{m}}$
Then amplitude A is,
$A={{A}_{0}}{{e}^{-kt}}$
substituting the values of amplitude, initial amplitude and time in the above equation we get,
0.9${A_0}$=${{A}_{0}}{{e}^{-5k}}$
$\Rightarrow$ 0.9=${{e}^{-5k}}$
Taking natural log on both sides,
$\Rightarrow$ ln(0.9)= -5k
At time t=15s equation becomes,
$A={{A}_{0}}{{e}^{-15k}}={{A}_{0}}{{e}^{3ln(0.9)}}$
= ${{A}_{0}}{{(0.9)}^{3}}=0.729{{A}_{_{0}}}$

Hence,option (B) is correct.