Question

For the damped oscillator shown, the mass m of the block is 200 g, k = 90 N m${}^{-1}$ and the damping constant b is 40 g s${}^{-1}$. What is the period of oscillation, time taken for its amplitude of vibrations to drop to half of its initial value.

• A. 1s, 6.93s
• B. 0.3s, 6.93s
• C. 0.6s, 6.93s
• D. 0.3s, 3s

Answer 1

The correct option is B

0.3s, 6.93s

Here km $=90\times 0.2=18k{g}^2{s}^{-2}$

$\Rightarrow \sqrt{km}=4.283kg{s}^{-1}$

Given value of $b=0.04kg{s}^{-1}$

We can see that $b<<\sqrt{km}$

So we can say the angular frequency of oscillation

$\omega \approx \sqrt{\frac{k}{m}}$

$\Rightarrow T\approx 2\pi \sqrt{\frac{m}{k}}=2\times \pi \times \sqrt{\frac{0.2}{90}}=0.3s$

The amplitude depends on time as

$A\left(t\right)=A_0{e}^{\frac{-bt}{2m}}$

So when amplitude drops to half it's initial value we have

$A\left(t\right)=\frac{A_0}{2}$ $\Rightarrow \frac{A_0}{2}=A_0{e}^{\frac{-bt}{2m}}$

$\Rightarrow \frac{1}{2}=e^{\frac{-bt}{2m}}$

$\Rightarrow -ln2=\frac{-bt}{2m}$

$\Rightarrow t=\frac{2mln2}{b}$

$=\frac{2\times 0.2\times ln2}{0.04}$

$=6.93s$