Question

1. The frequency of a damped harmonic oscillator is 100 Hz, and the ratio of the amplitude of two successive maxima is one half. a. What is the natural (undamped) frequency of this oscillator, in Hertz? b. If the oscillator is launched at time t0 from the origin with speed 2 m/s, what is its speed at time t 0.0140 sec?

Answer 1

Using a relation to find the value of _$\beta$ :

e^{-$\gamma$T_damped} = (1/2)

-_$\gamma$ T_damped = ln (0.5)

_$\gamma$ = (0.693) / T_damped

where, T_damped = time period of damped oscillator = 1 / f_damped = 1 / (100 Hz)

_$\gamma$ = (0.693) / [1 / (100 Hz)]

_$\gamma$ = 69.3 Hz

An angular frequency of a damped oscillator will be given by -

$\omega$_d = 2$\pi$ f_d

$\omega$_d = [(6.28 rad) (100 Hz)]

$\omega$_d = 628 Hz

The damped frequency is related to the natural frequency.

(a) The natural frequency of this oscillator which will be given by -

$\omega$₀ = _$\sqrt{}$$\omega$_d² + _$\gamma$²

$\omega$₀ = _$\sqrt{}$(628 Hz)² + (69.3 Hz)²

$\omega$₀ = _$\sqrt{}$399186.5 Hz²

$\omega$₀ = 631.8 Hz

$\omega$₀$\approx$ 632 Hz