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Wize University Physics Textbook (Master) > Periodic Motion: Oscillations

Underdamped Oscillations

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Damped Oscillation


In real systems, there are always resistive forces that oppose oscillation. These resistive forces cause the total energy of the system to decay. A type of these resistive forces is proportional to the speed of the oscillator and is show by:

Fresistive=−bvF_{resistive}=-bv

where bb is the damping constant and shows the strength of the resistive force.


The net force acting on the oscillator in this case is equal to:

F=−kx−bv=maF=-kx-bv=ma

The solution of the above equation for x(t)x\left(t\right)and the behaviour of the physical system depend on the magnitude of the damping constant.


Watch Out!
We still see some oscillations only for relatively small damping constants. This condition is known as underdamped condition and the oscillation in this case is known as damped oscillation.




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Underdamped Condition

For small values of damping constants(b2m<km)\left(\frac{b}{2m}<\sqrt{\frac{k}{m}}\right), the object undergoes an underdamped oscillation which is described by:


x(t)=Aoe−bt2mcos⁡(ω′t+φ)\boxed{x\left(t\right)=A_oe^{-\frac{bt}{2m}}\cos\left(\omega't+\varphi\right)}




Wize Concept
The system still oscillates in underdamped condition but with a decaying amplitude. The amplitude of this oscillation is given by:

A(t)=Aoe−bt2m\boxed{A\left(t\right)=A_oe^{-\dfrac{bt}{2m}}}

where AoA_ois the initial amplitude or A at  t=0.A\ \text{at} \ \ t=0.

Note: The larger the damping constant bb is, the faster AA goes to zero.


The Time Constant τ\tau is defined as τ=2mb\tau=\dfrac{2m}{b}and it is a measure of damping strength. It is a time it takes for the amplitude of the oscillation to drop to 1/e1/eof its initial value.

Wize Tip
The bigger the time constant is, the longer time it takes for the amplitude to decay to 1/e1/eof its initial value which means that there is a smaller damping constant.










Underdamped oscillation happens with a different angular velocity than the one for SHM. This angular frequency is known as damped angular frequency and is usually shown by ωdamped\omega_{damped},orω′\omega':


  ω′=ωo2−(b2m)2=km−(b2m)2  \boxed{\ \ \omega'=\sqrt{\omega_o^2-\left(\frac{b}{2m}\right)^2}=\sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^2}\ \ }
  • ωo\omega_ois the angular frequency of the undamped system (SHM) and is also known as natural frequency.















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Energy in a Damped Oscillation


Energy is no longer conserved in a damped oscillation. The total energy of the system could still be obtained by 12kA2\frac{1}{2}kA^2but here AA is a function of time:


E(t)=12kA2(t)=12kAo2e−btm\boxed{E\left(t\right)=\frac{1}{2}kA^2\left(t\right)=\frac{1}{2}kA_o^2e^{-\frac{bt}{m}}}





Watch Out!
Pay attention to the difference in the exponent of ee in E(t)E(t) and A(t)A(t)! The energy exponent is twice the amplitude exponent. This means that energy function decays faster than the amplitude function.












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Example: Decaying Amplitude vs Decaying Energy


A 0.3 kg0.3\ kgmass undergoes a damped oscillation with the damping constant 5 kgs5\ \frac{kg}{s}. How long does it take for the amplitude and total energy of the oscillation to be 14\frac{1}{4}of their initial values?

Solution:

Let's set t1t_1to be the time at which amplitude is dropped to 14\frac{1}{4}of its initial value.

A(t1)=A0 e−bt12m=14A0A\left(t_1\right)=A_{0\ }e^{-\dfrac{bt_1}{2m}}=\dfrac{1}{4}A_0→e−bt12m=14→ln⁡(14)=−bt12m=−ln⁡(4)\to e^{-\dfrac{bt_1}{2m}}=\dfrac{1}{4}\to\ln\left(\dfrac{1}{4}\right)=-\dfrac{bt_1}{2m}=-\ln\left(4\right)

→ln⁡(4)=bt12m→t1=2mln⁡(4) b=2×0.3×ln⁡(4)5=0.1663 s\to\ln\left(4\right)=\dfrac{bt_1}{2m}\to t_1=\dfrac{2m\ln\left(4\right)\ }{b}=\dfrac{2\times0.3\times\ln\left(4\right)}{5}=0.1663\ s

Now let's do the same calculations for energy and set t2t_2as the time at which energy is dropped to 14\frac{1}{4}of its initial value.

E(t2)=12kA02e−bt2m=14E0=14×12kA02E\left(t_2\right)=\dfrac{1}{2}kA_0^2e^{-\dfrac{bt_2}{m}}=\dfrac{1}{4}E_0=\dfrac{1}{4}\times\dfrac{1}{2}kA_0^2→e−bt2m=14→ln⁡(4)=bt2m\to e^{-\dfrac{bt_2}{m}}=\dfrac{1}{4}\to\ln\left(4\right)=\dfrac{bt_2}{m}

→t2=mln⁡(4)b=0.3ln⁡(4)5=0.0832 s\to t_2=\dfrac{m\ln\left(4\right)}{b}=\dfrac{0.3\ln\left(4\right)}{5}=0.0832\ s

As you can see it takes shorter for energy to decay to one quarter of its initial value compared to amplitude.

50 kgkg person taking a bungee jump:
If the initial length of the bungee cord is 20 mm and at the lowest point the length of the cord is 25mm, what is the spring constant of the cord?

Extra Practice
Simple harmonic motion: Pendulum
A large pendulum has 10 m10\ mlength and it has a mass of 50 kg50\ kg attached to that. The pendulum is oscillating in an under-damped condition. The initial amplitude of the oscillation is 1.2 m1.2\ mand it drops to 0.90 m0.90\ mafter 5 days. Calculate the time it takes for the total energy of this pendulum to drop by 110\frac{1}{10}factor (in days) . Include 3 decimal places in your answer.
Simple Harmonic Motion: Pendulum
A large pendulum has 10 m length and it has a mass of 50 kg attached to that. The pendulum is oscillating in an under-damped condition. The initial amplitude of the oscillation is 1.2 m and it drops to 0.90 m after 5 days. Calculate the time it takes for the total energy of this pendulum to drop by 1/10 factor.
Underdamped Oscillation
A mass-spring (m=1kg, k=10N/m, b=1 Kg/s) starts oscillation at time 0;
a) What is the frequency of oscillation?
b) By how much does the amplitude decrease after 3s?