You will work with a complex system to do the following:
Simple harmonic motion is motion which results from a Hooke's law force,
| $\textbf{F} = -k\textbf{x}$, | (1) |
and such motion is described by the function
| $x(t) = A\cos(\omega_0 t + \delta)$ | (2) |
where $A$ is the amplitude, $\delta$ is a phase, and $\omega_0$ is the natural frequency,
| $\omega_0 = \sqrt{k/m}$. | (3) |
Today we will investigate how the equations of motion change when we add additional forces – damping forces of the form $\textbf{F} = -b\textbf{v}$ and sinusoidal driving forces of the form $\textbf{F} = \textbf{F}_0 \cos \omega t$.
The oscillating system we will use today consists of a glider with attached springs on an air track (see Fig. 1). The air track provides a cushion of air which reduces friction on the cart, allowing us to more easily visualize the motion due to the other forces acting on the mass.
Let's discuss the more important parts of this system.
To help visualize and fit your data, we provide a Google Colab notebook.
In Part 1, you measured the mass, spring constant, and period of a single mass and single spring system. Today, you will repeat this procedure, but for the more complex system consisting of two springs attached to a glider (which is also attached to a string and hanging mass clip which move along with it).
Set the apparatus as shown in Fig. 1. Attach one spring to each end of the glider using the rubber band. Cut a string to about 1.5 meters in length and secure one end of the string in the lower end-hole of the glider using a small cork. Pass the string over the rotary encoder pulley and attach one small mass clip to the free end. Note where the glider is currently sitting. This will be the equilibrium position of the glider.
Turn on the blower to provide air to your air track. Open the SHM configuration file on the desktop and check to make sure the rotary encoder is working by recording some data as you move the cart around.
If the file isn't there, or if you need a fresh, unedited copy, it is available to download here: shm.cmbl.
Your tasks are the following:
Below is some advice that may be useful.
Mass
The “mass” which will be oscillating in this experiment is a metal glider plus the rubber band, the two magnets, the cork, the string, and the single mass clip attached to the string.
Spring constant
You can use a similar technique to determine the spring constant as you did last week: looking at the displacement of the mass as a function of force (due to hanging additional weight on the string).
Period
You will want to measure the period, $T$ (or equivalently, the angular frequency, $\omega = 2\pi/T$) for the system in two case: with both magnets are pointing up and with both magnets pointing down.
If, in addition to a Hooke’s Law restoring force, a damping force is also applied to the oscillating object, the resulting motion will change. Many damping forces may be represented by
| $\textbf{F} = -b\textbf{v}$, | (5) |
(i.e., by a force proportional to the velocity of the object, directed oppositely to its direction of motion) where $b$ is the damping constant. With damping included, the equation of motion is
| $m\dfrac{d^2x}{dt^2} + b\dfrac{dx}{dt} + kx = 0$, | (7a) |
which can be rewritten as
| $\dfrac{d^2x}{dt^2} + 2\xi\dfrac{dx}{dt} + \omega_0^2 x = 0$, | (7b) |
where we introduce the damping ratio,
| $\xi = \dfrac{b}{2m}$. | (8) |
This function has different solutions depending on the magnitude of $\xi$. Today, our system will be underdamped (meaning $\xi < \omega_0$). The solution to the equation in this case is given by
| $x(t) = A e^{-\xi t}\cos(\omega^{\prime} t + \delta)$, | (9) |
where $A$ is the initial amplitude of oscillation, $\delta$ is the initial phase of the oscillation, and $\omega^{\prime}$ is the new damped oscillation frequency,
| $\omega^{\prime} = \sqrt{\dfrac{k}{m} - \bigg(\dfrac{b}{2m}\bigg)^2} = \sqrt{\omega_0^2 - \xi^2}$. | (10) |
Hence, we see that the damped frequency is less than the natural frequency $\omega_0$.
A plot of Eq. (9) is shown in Fig. 2. (This should look familiar from Part 1! Your simple mass on a spring system was an underdamped system.)
In order to quantify our damped oscillatory motion, we want to make a few measurements.
First, with the magnets down (i.e. with damping), do the following:
*From the Analyze menu, select Curve Fit. In the General Equation list box, select Damped Harmonic. In the equation that appears, the meaning of the constants are as follows:
Now, repeat the above for the situation where the magnets facing up. This will remove the damping force due to the eddy currents, but there are still small frictional forces at play (so we aren't quite at a point of “zero damping”, but we can at least say we are “less damped” than before.)
Suppose now we add to the damped oscillator a periodic driving force of the form
| $\textbf{F} = \textbf{F}_0 \cos (\omega t)$, | (11) |
where $\omega$ is the frequency of the driving force and $\textbf{F}_0$ is the amplitude of the driving force.
Adding this force to the Hooke's law restoring force and the damping force we find a total equation of motion
| $m\dfrac{d^2x}{dt^2} + b\dfrac{dx}{dt} + kx = F_0 \cos(\omega t)$ | (12a) |
or
| $\dfrac{d^2x}{dt^2} + 2\xi\dfrac{dx}{dt} + \omega_0^2 x = \dfrac{F_0}{m}\cos(\omega t)$. | (12b) |
The solution to this differential equation is complicated at early times, but if we wait long enough for the transient behavior to die out, then the long time steady solution is
| $x(t) = A_D \cos(\omega t + \phi)$, (for $t \gg 0$) | (13) |
where the amplitude $A_D$ is given by
| $A_D = \dfrac{F_0/m^2}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\xi^2\omega^2}}$ | (14) |
and the phase is given by
| $\phi = \tan^{-1}\bigg[\dfrac{2\xi\omega}{(\omega^2 - \omega_0^2)}\bigg]$. | (15) |
Equation (14) tells us that for constant amplitude driving force $F_0$, the amplitude of the oscillation is a function of $\omega$. In particular, it tells us that the amplitude of oscillation has a maximum when $\omega = \omega_0$, i.e., when the driving frequency is in resonance with the natural frequency. It also tells us that if there were no dissipative forces ($b=0$, or equivalently, $\xi = 0$), then the amplitude would go to infinity at the resonant frequency no matter how small the driving force!
A plot of Eq. (14) is shown in Fig. 3 for different values of damping coefficient, $\xi$. As should be apparent from this figure, the amplitude of the oscillation at resonance is higher than at any other frequency, and the magnitude of the resonant oscillation gets larger as $\xi$ gets smaller.
The phase angle $\phi$, Eq. (15), specifies the time relationship between the driving force and the resultant displacement as a function of $\xi$, the driving frequency $\omega$, and the natural frequency $\omega_0$. A plot of phase angle versus driving frequency is shown in Fig. 4.
We see the following from the figure:
For the following measurements, we shall use the magnets, mounted on the glider mass pins, with their pole faces pointing down toward the air track.
Turn on the motor and observe the behavior of the glider as you adjust the frequency.
Re-orient the magnets with their pole faces pointing away from the track.
Tune the driving frequency for maximum amplitude of the glider.
NOTE: You do not need to measure and plot the full resonance curve for the undamped case (though there is a place within the Google Colab to do so if you want!)
The quality factor, $Q$, of an oscillating system is a measure of how many oscillations will take place before the motion dies out. There are several ways to express the $Q$ of an oscillating system.
In terms of energy, we can define $Q$ as
| $Q = 2\pi \dfrac{\textrm{energy stored in the system}}{\textrm{energy lost per cycle}}$. |
In terms of amplitude, we can define $Q$ as
| $Q = \dfrac{\textrm{amplitude at resonance}}{\textrm{amplitude at small driving frequency}}$. |
But, the definition we will use for our system today is the definition revolving around the width of the resonance peak:
| $Q = \dfrac{\omega_0}{\Delta \omega}$ |
where $\Delta \omega$ is the full width at half maximum (FWHM) of the peak (as shown in Fig. 4).
|
| Figure 4: The full width at half maximum of a curve is the distance (along the $x$-axis) between the two points where the amplitude is half the maximum value (along the $y$-axis). (Source: Wikipedia) |
The full width at half maximum is related to the damping factor as
| $\Delta \omega = 2\xi$ |
such that
| $Q = \dfrac{\omega_0}{2\xi}$. |
Make sure to submit your lab notebook by the end of the period. Download a copy of your notebook in PDF format and upload it to the appropriate spot on Canvas.
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