- #1
zenterix
- 605
- 79
- Homework Statement
- Using a force of ##4\text{N}##, a damped harmonic oscillator is displaced from equilibrium by ##0.2\text{m}##.
At ##t=0## it is released from rest. The resultant displacement of the oscillator from equilibrium as a function of time is displayed below.
- Relevant Equations
- Using the given information, estimate as best you can
a) mass of oscillator
b) quality factor of oscillator
The equation of motion is
$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=0\tag{1}$$
The roots of the characteristic polynomial are
$$\alpha=-\frac{\gamma}{2}\pm\sqrt{\frac{\gamma^2}{4}-\omega_0^2}\tag{2}$$
and since we have oscillations we are in the underdamped case in which
$$\gamma^2<\frac{\omega_0^2}{4}\tag{2a}$$
The solution to (1) is
$$x(t)=e^{-\frac{\gamma t}{2}}(c_1\cos{\omega t}+c_2\sin{\omega t})\tag{3}$$
where
$$\omega=\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{4}$$
The velocity is
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-c_1\omega\sin{\omega t}+c_2\omega\cos{\omega t})\tag{5}$$
Using the initial conditions ##x(0)=0.2## and ##\dot{x}(t)=0## we have
$$x(0)=c_1=0.2\tag{6}$$
$$\dot{x}(0)=-\frac{\gamma}{2}\cdot 0.2+c_2\omega=0\tag{7}$$
$$\implies c_2=0.1\frac{\gamma}{\omega}\tag{8}$$
Thus
$$x(t)=e^{-\frac{\gamma t}{2}}(0.2\cos{\omega t}+0.1\frac{\gamma}{\omega}\sin{\omega t})\tag{9}$$
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-0.2\omega\sin{\omega t}+0.1\gamma\cos{\omega t})\tag{10}$$
At this point, we've determined what the motion is given some unknown parameters.
Assuming the underdamping condition (2a) is true then each different choice of ##m, \gamma##, and ##k## determines the angular frequency of the resulting oscillations.
The forms of (9) and (10) don't make the amplitude and phase of the oscillations very explicit. These variable are, however, also a function of our choice of parameters.
We can see this by writing (3) in amplitude-phase form
$$x(t)=e^{-\frac{\gamma t}{2}} A\cos{(\omega t-\phi)}\tag{11}$$
$$A=\sqrt{c_1^2+c_2^2}=\frac{1}{10}\sqrt{4+\frac{\gamma^2}{\omega^2}}\tag{12}$$
$$\phi=\tan^{-1}{\frac{c_2}{c_1}}=\tan^{-1}{\frac{\gamma}{2\omega}}\tag{13}$$
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)-e^{-\frac{\gamma t}{2}}A\omega\sin{(\omega t-\phi)}\tag{14}$$
At this point, we look at the given plot
My question is about estimating the parameters from this plot.
Recall that what we want to determine ultimately is the mass that is oscillating and the quality factor of the oscillator.
The latter is defined by
$$Q=\frac{1}{\gamma}\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{15}$$
Here is what I did.
The first thing I did was to determine the spring constant from the original differential equation.
$$\ddot{x}(0)+\gamma\dot{x}(0)+\omega_0^2x(0)=0\tag{16}$$
$$\frac{F_s}{m}+\omega_0^2\cdot 0.2=0\tag{17}$$
where at ##t=0## the magnitude of the spring force equals the force we were applying to hold the spring in its initial position, and the sign is opposite.
$$\frac{-4}{m}+\frac{k}{m}\cdot 0.2=0\tag{18}$$
$$\implies k=20\mathrm{\frac{N}{m}}\tag{19}$$
In words, we determined the spring constant by noting that at position ##0.2## the spring exerts a force of ##-4\text{N}## and this force has the formula ##F_s=-kx##. Thus, ##k=\frac{F_s}{x}##.
Suppose we try to estimate the pseudo-period from the plot.
I've drawn in some dashed lines in the plot to try to do this
The pseudo-period seems to be slightly larger than 3. Perhaps 3.2.
Before I continue, I have a question.
Since the mass is released from rest, can we assume that the phase is zero?
From (13) it seems we cannot as this would imply that ##\gamma=0##. But if we are holding the mass and releasing, doesn't it start at the maximum of a pseudo-oscillation? Ie, at the top of the damped cosine? Is it not then in phase with cosine?
$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=0\tag{1}$$
The roots of the characteristic polynomial are
$$\alpha=-\frac{\gamma}{2}\pm\sqrt{\frac{\gamma^2}{4}-\omega_0^2}\tag{2}$$
and since we have oscillations we are in the underdamped case in which
$$\gamma^2<\frac{\omega_0^2}{4}\tag{2a}$$
The solution to (1) is
$$x(t)=e^{-\frac{\gamma t}{2}}(c_1\cos{\omega t}+c_2\sin{\omega t})\tag{3}$$
where
$$\omega=\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{4}$$
The velocity is
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-c_1\omega\sin{\omega t}+c_2\omega\cos{\omega t})\tag{5}$$
Using the initial conditions ##x(0)=0.2## and ##\dot{x}(t)=0## we have
$$x(0)=c_1=0.2\tag{6}$$
$$\dot{x}(0)=-\frac{\gamma}{2}\cdot 0.2+c_2\omega=0\tag{7}$$
$$\implies c_2=0.1\frac{\gamma}{\omega}\tag{8}$$
Thus
$$x(t)=e^{-\frac{\gamma t}{2}}(0.2\cos{\omega t}+0.1\frac{\gamma}{\omega}\sin{\omega t})\tag{9}$$
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)+e^{-\frac{\gamma t}{2}}(-0.2\omega\sin{\omega t}+0.1\gamma\cos{\omega t})\tag{10}$$
At this point, we've determined what the motion is given some unknown parameters.
Assuming the underdamping condition (2a) is true then each different choice of ##m, \gamma##, and ##k## determines the angular frequency of the resulting oscillations.
The forms of (9) and (10) don't make the amplitude and phase of the oscillations very explicit. These variable are, however, also a function of our choice of parameters.
We can see this by writing (3) in amplitude-phase form
$$x(t)=e^{-\frac{\gamma t}{2}} A\cos{(\omega t-\phi)}\tag{11}$$
$$A=\sqrt{c_1^2+c_2^2}=\frac{1}{10}\sqrt{4+\frac{\gamma^2}{\omega^2}}\tag{12}$$
$$\phi=\tan^{-1}{\frac{c_2}{c_1}}=\tan^{-1}{\frac{\gamma}{2\omega}}\tag{13}$$
$$\dot{x}(t)=-\frac{\gamma}{2}x(t)-e^{-\frac{\gamma t}{2}}A\omega\sin{(\omega t-\phi)}\tag{14}$$
At this point, we look at the given plot
My question is about estimating the parameters from this plot.
Recall that what we want to determine ultimately is the mass that is oscillating and the quality factor of the oscillator.
The latter is defined by
$$Q=\frac{1}{\gamma}\sqrt{\omega_0^2-\frac{\gamma^2}{4}}\tag{15}$$
Here is what I did.
The first thing I did was to determine the spring constant from the original differential equation.
$$\ddot{x}(0)+\gamma\dot{x}(0)+\omega_0^2x(0)=0\tag{16}$$
$$\frac{F_s}{m}+\omega_0^2\cdot 0.2=0\tag{17}$$
where at ##t=0## the magnitude of the spring force equals the force we were applying to hold the spring in its initial position, and the sign is opposite.
$$\frac{-4}{m}+\frac{k}{m}\cdot 0.2=0\tag{18}$$
$$\implies k=20\mathrm{\frac{N}{m}}\tag{19}$$
In words, we determined the spring constant by noting that at position ##0.2## the spring exerts a force of ##-4\text{N}## and this force has the formula ##F_s=-kx##. Thus, ##k=\frac{F_s}{x}##.
Suppose we try to estimate the pseudo-period from the plot.
I've drawn in some dashed lines in the plot to try to do this
The pseudo-period seems to be slightly larger than 3. Perhaps 3.2.
Before I continue, I have a question.
Since the mass is released from rest, can we assume that the phase is zero?
From (13) it seems we cannot as this would imply that ##\gamma=0##. But if we are holding the mass and releasing, doesn't it start at the maximum of a pseudo-oscillation? Ie, at the top of the damped cosine? Is it not then in phase with cosine?