Question

Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass $=500\text{g}$, damping constant $=20\text{g/s}$ then how much time is required for the amplitude of the system to drop to half of its initial value?
$(ln2=0.693)$

• A. $15.01\text{s}$
• B. $17.32\text{s}$
• C. $0.034\text{s}$
• D. $34.65\text{s}$

Answer 1

The correct option is D $34.65\text{s}$

In damped oscillation, amplitude is given as: $$A=A_0{e}^{-\frac{bt}{2m}}$$ From question, $A=\frac{A_0}{2}$
$\Rightarrow \frac{1}{2}=e^{-\frac{bt}{2m}}$

$\Rightarrow \frac{bt}{2m}=ln2=0.693$

$\Rightarrow t=\frac{2m}{b}\times 0.693$

$\Rightarrow t=2\times \frac{500}{20}\times 0.693$

$\therefore t=50\times 0.693=34.6\text{s}$