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# The displacement of a damped harmonic oscillator is given by $x\left( t \right) = {e^{ - 01.1t}}\cos \left( {10\pi t + \phi } \right)$. Here $t$ is in seconds. The time taken for its amplitude of vibration to drop to half for its initial value is close to:(A) $13\,\sec$(B) $7\,\sec$(C) $27\,\sec$(D) $4\,\sec$

Last updated date: 15th Jul 2024
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Hint The time taken for its amplitude of vibration to drop to half for its initial value is close to can be determined by using the formula of the simple harmonic equation of the time, then the time taken for its amplitude of vibration can be determined.

Useful formula
The formula of the simple harmonic equation of the time,
$A = {A_0}{e^{ - kt}}$
Where, $A$ is the amount of the vibration, ${A_0}$ is the initial amount of the vibration and $t$ is the time taken.

Complete step by step solution
Given that,
The displacement of a damped harmonic oscillator is given by, $x\left( t \right) = {e^{ - 01.1t}}\cos \left( {10\pi t + \phi } \right)$.
From the given equation, then the value of the $k$ is $0.1$,
Now,
The formula of the simple harmonic equation of the time,
$A = {A_0}{e^{ - kt}}$
By substituting the value of the $k$ in the above equation, then the above equation is written as,
$A = {A_0}{e^{ - 0.1t}}$
Now, the above equation is also written as,
$A = {A_0}{e^{ - 0.1t}} = \dfrac{{{A_0}}}{2}$
By cancelling the same terms in the above equation, then the above equation is written as,
${e^{ - 0.1t}} = \dfrac{1}{2}$
To remove the exponential term in the LHS of the above equation, then the above equation is written as,
$\ln 2 = 0.1t$
By taking the $\ln$ in the LHS of the above equation, then the above equation is written as,
$0.693 = 0.1t$
By rearranging the terms in the above equation, then the above equation is written as,
$t = \dfrac{{0.693}}{{0.1}}$
By dividing the terms in the above equation, then the above equation is written as,
$t = 6.93\,\sec$
Then the above equation is also written as,
$t = 7\,\sec$

Hence, the option (B) is the correct answer.

Note In physical calculation, there are two types of the logarithmic, one type of logarithmic is $\log$ and the other type of the logarithmic is $\ln$. So, during the calculation, the students must give concentration in the logarithmic step. To remove the exponential term, $\ln$ is used.