Answer to Question #107290 in Mechanics | Relativity for Parul

Question

For a damped harmonic oscillation the equation of motion is md2x/dt2 + ¥dx/dt+ kx=0 with m= 0.20kg ¥ = 0.04kg/s and k= 65N/m .
Calculate i) the period of motion
ii) number of oscillation in which it's amplitude will become half of it's initial value and iii) the number of oscillation in which it's mechanical energy will drop to half of it's initial value

Answer 1

$m{\frac {d^2x} {dt^2}}+\gamma{\frac {dx} {dt}}+kx=0$

$m=0.2kg$

$\gamma=0.04kg/s$

$k=65N/m$

i) $T={\frac {2\pi} {\sqrt{{\frac k m}-{\frac {\gamma^2} {4m^2}}}}}=0.35s$

ii) Amplitude $A=A_0e^{-\beta t}$ , where $\beta=\gamma/2m$

number of oscillation is $n=t/T$ in which it's amplitude will become half of it's initial value can be found as this $A_0/2=A_0e^{-\beta nT}$

$\beta nT=ln(2)$

$n={\frac {2mln(2)} {\gamma T}}=19.8$ approx to 19

iii) mechanical energy is ${\frac {kA^2} 2}$. number of oscillation is n=t/T

n=t/T in which mechanical energy will drop to half of it's initial value can be found as this

${\frac {kA^2} 2}={\frac {kA_0^2} 2}e^{-2\beta t}$

$\beta nT=ln(\sqrt2)$

$n={\frac {2mln(\sqrt2)} {\gamma T}}=9.9$ approx to 9