Question

An undamped oscillator has period ro = 1.000 s, but I now add a little damping so that its period changes to r i = 1.001 s. What is the damping factor 8? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

Answer 1

Answer:

Step-by-step explanation:

a) Under the undamped oscillator, the equation that relates frequencies is:

$\omega_{1}=\sqrt{\omega^{2}_{0}-\beta^{2}}$            (1)

when β is the damping factor and ω₀ is the natural frequency.

Let's solve the equation (1) for β.

$\beta=\sqrt{\omega^{2}_{0}-\omega^{2}_{1}}=\omega_{0}\sqrt{1-\frac{\omega^{2}_{1}}{\omega^{2}_{0}}}=\omega_{0}\sqrt{1-\frac{T^{2}_{0}}{T^{2}_{1}}}$.

So  β = 0.2808 1/s.

b) Now, the amplitud equation is: $A=e^{\beta t }$, after 10 cycles we will have t=10*T₁ and $A=e^{\beta 10 \cdot T_{1}} = e^{0,2802*10*1.001} = 0.061$.

In says that after 10 cycles the amplitude of oscillation decrease by a factor of 0.06 or about 6%. In the case of the period, we have 0.1%, concluding that the change of amplitude is more noticeable.  

Have a nice day!