The equation of motion of a harmonic oscillator is given by$\frac{{{d^2}x}}{{d{t^2}}} + 2\zeta {\omega _n}\frac{{dx}}{{dt}} + \omega _n^2x = 0$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

Question

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

$X$

1.	The equation of motion of a harmonic oscillator is given by\(\frac{{{d^2}x}}{{d{t^2}}} + 2\zeta {\omega _n}\frac{{dx}}{{dt}} + \omega _n^2x = 0\)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» B. \(X{e^{2n\pi \left( {\frac{\zeta }{{\sqrt {1 - {\zeta ^2}} }}} \right)}}\)

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