Question

The equation of motion of ad amped harmonic oscillator is given by
d 2 x + 2 b d x + ω 02 x = 0 dt2 dt
with m = 0.25kg, b = 0.14s−1 and ω0 =18.4s−1.
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 # 76071, Physics -Mechanics- Relativity:

Question: The equation of motion of a damped harmonic oscillator is given by

$$\frac {\mathrm {d} ^ {2} \mathbf {x}}{\mathrm {d t} ^ {2}} + 2 \mathbf {b} \frac {\mathrm {d} \mathbf {x}}{\mathrm {d t}} + \omega_ {0} ^ {2} \mathbf {x} = 0$$

with $m = 0.25\mathrm{kg}$ , $b = 0.14\mathrm{sec}^{-1}$ and $\omega_0 = 18.4\mathrm{sec}^{-1}$ .

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.

Solution: The equation of motion of a damped harmonic oscillator is given by

$$\frac {\mathrm {d} ^ {2} \mathbf {x}}{\mathrm {d t} ^ {2}} + 2 \mathbf {b} \frac {\mathrm {d} \mathbf {x}}{\mathrm {d t}} + \omega_ {0} ^ {2} \mathbf {x} = 0$$

Given, mass $(\mathfrak{m}) = 0.25\mathrm{kg}$

Damping constant $(b) = 0.14\mathrm{sec}^{-1}$

Damping frequency $(\omega_0) = 18.4\mathrm{sec}^{-1}$

(i). solution of the equation (1) is $x(t) = A \exp(-bt)$ . Cos $\omega_0 t$ ...(2)

Or, $x(t) = A \exp(-0.14t)$ . Cos $\omega_0 t$

Now amplitude $= A \exp(-0.14t)$

So, time period $= \frac{2\pi}{\omega_0} = \frac{2\pi}{18.4} = 0.341$ sec.

(ii). Amplitude becomes half of its initial value i.e. $\exp(-0.14t) = \frac{1}{2}$ or, $t = 4.95$ sec.

So, the number of oscillation $= \frac{4.95}{0.341} = 14.52$

(iii). Mechanical energy of the oscillator is $\frac{1}{2} x \omega_0 x$ (amplitude) $^2$ = $\frac{1}{2} x 18.4 x(A)^2 x \exp(-0.28t)$

Now mechanical energy becomes half of its initial value i.e. $\exp(-0.28t) = \frac{1}{2}$ or, $t = 2.47$ sec.

So, the number of oscillation $= \frac{2.47}{0.341} = 7.24$

Answer: (i). 0.341 sec.

(ii). 14.52

(iii). 7.24

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