Question

The equation of motion of a damped harmonic oscillator is given by
d2x/dt2+2b dx/dt + w2x=0. m= 0.25 kg , b= 0.14/s and w = 18.4/s Calculate i) the time period; ii) number of oscillations in which its amplitude will become half of its initial value; and iii) number of oscillations in which its mechanical energy will reduce to half of its initial value.

Answer 1

Answer on Question #73473, Physics / Mechanics | Relativity

Question. The equation of motion of a damped harmonic oscillator is given by $\frac{d^2x}{dt^2} + 2b\frac{dx}{dt} + \omega_0^2 x = 0$, $m = 0.25\,kg$, $b = 0.14\frac{1}{s}$ and $\omega_0 = 18.4\frac{1}{s}$ Calculate

i) the time period;

ii) number of oscillations in which its amplitude will become half of its initial value;

iii) number of oscillations in which its mechanical energy will reduce to half of its initial value.

Solution.

i) If $\omega_0 = 18.4\frac{1}{s}$ then

$$\omega = 2\pi\nu = \frac{2\pi}{T} \rightarrow T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\omega_0^2 - \beta^2}} = \frac{2 \cdot 3.14}{\sqrt{18.4^2 - 0.14^2}} = 0.341\,s.$$

ii) The amplitude of a damped harmonic oscillator is

$$A = A_0 e^{-bt}.$$

So

$$\frac{A_0}{2} = A_0 e^{-0.14t} \rightarrow \frac{1}{2} = e^{-0.14t} \rightarrow t = -\frac{\ln\frac{1}{2}}{0.14} = 4.951\,s;$$ $$N_1 = \frac{t}{T} = \frac{4.951}{0.341} \approx 14.$$

iii) Mechanical energy of oscillations

$$E = \frac{m\omega^2 A^2}{2}.$$

We have

$$E_0 = \frac{m\omega^2 A_0^2}{2};$$ $$\frac{E_0}{2} = \frac{m\omega^2 A^2}{2};$$ $$\frac {E _ {0}}{\frac {E _ {0}}{2}} = \frac {m \omega^ {2} A _ {0} ^ {2}}{2} \cdot \frac {2}{m \omega^ {2} A ^ {2}} \rightarrow 2 = \frac {A _ {0} ^ {2}}{A ^ {2}} \rightarrow A = \frac {A _ {0}}{\sqrt {2}}.$$

Hence (see ii))

$$t = - \frac {\ln \frac {1}{\sqrt {2}}}{0 . 1 4} \approx 2. 4 7 s;$$ $$N _ {2} = \frac {t}{T} = \frac {2 . 4 7}{0 . 3 4 1} \approx 7.$$

Answer. $T = 0.341$ s; $N_{1} = 14$ ; $N_{2} = 7$ .

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