Question

For a damped harmonic oscillator of mass $200\, g$, the values of spring constant and damping constant are, respectively, $90\, N/m$ and $0.04\, kg/s$. The time taken for its amplitude of vibration of drop to half of its initial value is $(\log, 2 = 0.693)$

• A. 7.0 s
• B. 14.2 s
• C. 15.9 s
• D. 26.6 s

Answer 1

Answer: (A)

The given, $K=90\, N / m$
$b=0.04 \,kg / s$
$\log _{e}=2=0.693$
$m=200\, g =0.2\, kg$
The (instantaneous) amplitude of damped oscillations is given by
$A_{t}=A e^{-b t / 2\, m}$
Let $t_{1}$ be the time taken for the amplitude to drop to half of its initial value, that is at $t=t_{1}$
$A_{t}=A / 2$
Thus, $\frac{A}{2}=A e^{-b t_{1} / 2 m}$
$e^{b t_{1} / 2 m}=2$
$\frac{b t_{1}}{2 m}=\log _{e} 2$
$t_{1}=\frac{2 m}{b} \log _{e} 2$
$=\frac{2 \times 0.2}{0.04} \log _{e} 2=\frac{0.4}{0.04} \times 0.693$
$=10 \times 0.693=6.93 \,s$
or $\cong 7.0 \,s$