Question

The equation of motion of a harmonic oscillator is given by
$\frac{d^{2}x}{d{t}^2}+2\zeta {\omega }_n\frac{dx}{dt}+{\omega }_n^{2}x=0$
and the initial conditions at t = 0 are $x\left(0\right)=X$,
$\frac{dx}{dt}\left(0\right)=0.$ The amplitude of x(t) after n complete cycles is

• A. $X{e}^{-2n\pi \left(\frac{\zeta }{\sqrt{1-{\zeta }^2}}\right)}$
• B. $X{e}^{2n\pi \left(\frac{\zeta }{\sqrt{1-{\zeta }^2}}\right)}$
• C. $X{e}^{-2n\pi \left(\frac{\sqrt{1-{\zeta }^2}}{\zeta }\right)}$
• D. X

Answer 1

The correct option is A $X{e}^{-2n\pi \left(\frac{\zeta }{\sqrt{1-{\zeta }^2}}\right)}$

$X_1=X{e}^{-\zeta {\omega }_n{T}_d}$

After n complete cycles,

$X_n=X{e}^{-\zeta {\omega }_n{T}_d.n}\text{...(i)}$

We know that,

$T_d=\frac{2\pi }{{\omega }_d}=\frac{2\pi }{\sqrt{(1-{\zeta }^2)}{\omega }_n}$

Now putting the value of $T_d$ in equation (i)

$\therefore$ $X_n=X{e}^{-\zeta {\omega }_n.n.\frac{2\pi }{\sqrt{(1-{\zeta }^2}){\omega }_n}}$

$=X{e}^{-\zeta .n.\frac{2\pi }{\sqrt{(1-{\zeta }^2})}}$

$=X{e}^{-2n\pi \left(\frac{\zeta }{\sqrt{1-{\zeta }^2}}\right)}$