Question

For the damped oscillator system shown with m = 250g, k = 85 N/m, and b = 70 g/s, what is the ratio of the oscillation amplitude at the end of 20 cycles to the initial oscillation amplitude?

• A. 1:2
• B. 20:21
• C. 19:50
• D. 5:6

Answer 1

The correct option is C

19:50

The angular frequency of oscillation in this case will be equal to

$\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

$\omega =\sqrt{\frac{85}{0.25}-\frac{(0.07{)}^2}{4\times (0.25{)}^2}}\approx \sqrt{340}=18.4rad{s}^{-1}$

$T=\frac{2\pi }{\omega }=0.34s$

In 20 cycles 20 time periods must have passed, so time after 20 cycles, $t=20\times 0.34$

=6.8 s

We know $A=A_0{e}^{\frac{-bt}{2m}}$

$A=A_0{e}^{-\left(\frac{0.07\times 6.8}{2\times 0.25}\right)}=0.38A_0$

So the ratio $A:A_0=0.38:1or19:50$