An oscillator with a mass of $500g$and a period of $0.50s$has

an amplitude that decreases by $2.0\%$during each complete oscillation.

If the initial amplitude is $10cm$, what will be the amplitude

after $25$oscillations?

Damping oscillation: The mechanical energy $E$in a real oscillating system decreases during oscillation because external forces, such as resistance, prevent the oscillation and convert the mechanical energy into thermal energy. The real oscillator and its movement are then said to be damped. If the damping force gives $F=bv$, where v is the speed of the oscillation and b is the damping constant, then the displacement of the oscillation is given by

$x\left(t\right)=A{e}^{-bt/2m}cos(\omega t+\varphi )$

where $\omega$, the angular frequency of the damped oscillator can be given by

$\omega =\surd (g/L-b^2/(4m^2\left)\right)$

now$\surd (g/L)$is the angular frequency of an undamped oscillator $(b=0)$

• The mass of the oscillator can be:$m=\left(500\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.500\mathrm{kg}$.
• The time of the oscillator is: $T=0.50s$.
• The amplitude of the oscillator decreases $2.0\%$by during each complete oscillation.
• The initial amplitude of the oscillator can be$A_0=10$.

The objective is to find out the amplitude of the oscillator after$25$oscillations

Since, The amplitude of the oscillation decreases $2.0\%$by after a complete oscillation, the amplitude $A_1$after the first oscillation is

$A_1=(100\mathrm{\%}-2\mathrm{\%})A_0$

$=0.98A_0$

where $A_0$is the initial amplitude of the oscillation. The amplitude after second oscillation is $0.98$

$A_1:A_2=0.98\left(0.98A_0\right)={0.98}^2{A}_0$

The amplitude of the oscillator after $25$oscillations is

$A_{25}=(0.98{)}^{25}{A}_0$

$A_{25}=\left(0.98{)}^{25}\right(10\mathrm{cm})$

$=6.0\mathrm{cm}$