- #1
zenterix
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- 84
- Homework Statement
- Suppose we have a mass ##m## moving on a horizontal air track.
The mass is attached on both sides to two identical springs each with spring constant ##k## and relaxed length ##l_0##.
The end of the first spring is fixed.
The end of the second spring is attached to an electrical motor that causes it to undergo harmonic motion with amplitude ##d\sin{\omega_d t}##.
At ##t=0## the springs are relaxed and the mass is not moving.
At ##t=0## the motor is switched on.
Define the coordinate system such that the origin is at the position of the mass at ##t=0##.
- Relevant Equations
- Assume there is a small air friction given by ##b\dot{x}##.
The mass is under the effect of two forces ##F_a## and ##F_b##.
Let's denote the right edge of the mass as ##b## and the left edge as ##a##. Let's denote the right end of the right spring as ##c##.
Denote the width of the mass by ##w##.
The left hand side
##x## is the position of the middle of the mass.
##\Delta x_a## is the change in the position of ##a## and ##\delta x## is the change in the position of ##c##.
Then the positions are
$$r_a=-w+\Delta x_a\tag{1}$$
$$r_b=w+\Delta x_a\tag{2}$$
$$r_c=w+l_0+\Delta x_c\tag{3}$$
Any time the right spring is in equilibrium we have ##r_b-r_c=-l_0##.
$$(r_b-r_c)_{eq}=-l_0\tag{4}$$
In addition we have
$$\Delta x_a=x\tag{5}$$
$$\Delta x_c=d \sin{\omega_d t}\tag{6}$$
Thus
$$F_b=-k((r_b-r_c)-(r_b-r_c)_eq)\tag{7}$$
$$=k(\Delta x_c-\Delta x_a)\tag{8}$$
$$F_a=-k((r_a-r_{wall})-(r_a-r_{wall})_{eq}\tag{9}$$
$$=-k\Delta x_a\tag{10}$$
Equation of Motion
$$F_a+F_b+F_{drag}=m\ddot{x}\tag{11}$$
$$-k\Delta x_a+k(\Delta x_c-\Delta x_a)-b\dot{x}=m\ddot{x}\tag{12}$$
$$-kx+k(d\sin{\omega_d t}-x)-b\dot{x}=m\ddot{x}\tag{13}$$
$$\ddot{x}+\frac{b}{m}\dot{x}+\frac{2k}{m}x=\frac{k}{m}d\sin{\omega_d t}\tag{14}$$
$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=A_0\sin{\omega_d t}\tag{15}$$
Solution
$$x(t)=\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}\sin{(\omega_d t-\beta)}\tag{16}$$
with
$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)\tag{17}$$
One question I have is about writing the following
$$\tan{\beta}=\frac{\gamma\omega_d}{\omega_0^2-\omega_d^2}\tag{18}$$
If you try to plot this with Maple, you will get a result that isn't correct for the problem. It seems we lose information in going from (17) to (18).
What justifies writing (18) instead of (17)? If I recall correctly, I see (18) much more frequently than (17) in textbooks/notes/etc.
Numerical Example Parameters
Let's consider a concrete example.
Assume
$$m=1$$
$$k=1$$
$$d=1$$
$$\gamma=\frac{\omega_0}{4}$$
Then
$$A_0=\frac{kd}{m}=1$$
$$\omega_0=\sqrt{\frac{2k}{m}}=\sqrt{2}$$
$$\gamma=\frac{1}{4}$$
$$b=m\gamma=\frac{1}{4}$$
The steady-state solution is
$$x(t)=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}\sin{(\omega_d t-\beta)}$$
The amplitude of the steady-state solution is
$$B=\frac{A_0}{(\omega_0^2-\omega_d^2)^2-\gamma^2\omega_d^2}=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}$$
A plot of this function is
The phase lag of the steady-state solution is
$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)=arg\left (2-\omega_d^2+i\frac{\omega_d^2}{16}\right )$$
The amplitude maximum does not occur at ##\omega_0##. It occurs at
$$\omega_{d,max}=\sqrt{\frac{2\omega_0^2-\gamma^2}{2}}<\omega_0$$
My question is about the consideration of the initial conditions.
Suppose
$$\omega_d=\frac{\omega_0}{2}$$
Then, the steady-state solution is
$$x_p(t)=\frac{32}{73}\sin{(0.7t-0.34)}$$
The amplitude is ##32/73## and the phase lag is ##-0.34\text{rad}##.
I am interested in what is happening at ##t=0## here.
What is the meaning of the negative value of ##x_p(0)## here?
I guess that for the initial conditions implicit above, the homogeneous solution is just zero. But what does this mean exactly?
I think what I am trying to understand is the following.
First of all, is there more than one possible particular solution?
I know that the homogeneous solution has two parameters so there is an infinite number of those solutions.
Given a particular solution, and given initial conditions, we obtain the two parameters to pin down a homogeneous solution such that the general solution describes the motion of our mass with the initial conditions.
If we were to have started with a different particular solution, we would get other values for the homogeneous parameters (given the initial conditions).
Are there other particular solutions?
Let's denote the right edge of the mass as ##b## and the left edge as ##a##. Let's denote the right end of the right spring as ##c##.
Denote the width of the mass by ##w##.
The left hand side
##x## is the position of the middle of the mass.
##\Delta x_a## is the change in the position of ##a## and ##\delta x## is the change in the position of ##c##.
Then the positions are
$$r_a=-w+\Delta x_a\tag{1}$$
$$r_b=w+\Delta x_a\tag{2}$$
$$r_c=w+l_0+\Delta x_c\tag{3}$$
Any time the right spring is in equilibrium we have ##r_b-r_c=-l_0##.
$$(r_b-r_c)_{eq}=-l_0\tag{4}$$
In addition we have
$$\Delta x_a=x\tag{5}$$
$$\Delta x_c=d \sin{\omega_d t}\tag{6}$$
Thus
$$F_b=-k((r_b-r_c)-(r_b-r_c)_eq)\tag{7}$$
$$=k(\Delta x_c-\Delta x_a)\tag{8}$$
$$F_a=-k((r_a-r_{wall})-(r_a-r_{wall})_{eq}\tag{9}$$
$$=-k\Delta x_a\tag{10}$$
Equation of Motion
$$F_a+F_b+F_{drag}=m\ddot{x}\tag{11}$$
$$-k\Delta x_a+k(\Delta x_c-\Delta x_a)-b\dot{x}=m\ddot{x}\tag{12}$$
$$-kx+k(d\sin{\omega_d t}-x)-b\dot{x}=m\ddot{x}\tag{13}$$
$$\ddot{x}+\frac{b}{m}\dot{x}+\frac{2k}{m}x=\frac{k}{m}d\sin{\omega_d t}\tag{14}$$
$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=A_0\sin{\omega_d t}\tag{15}$$
Solution
$$x(t)=\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}\sin{(\omega_d t-\beta)}\tag{16}$$
with
$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)\tag{17}$$
One question I have is about writing the following
$$\tan{\beta}=\frac{\gamma\omega_d}{\omega_0^2-\omega_d^2}\tag{18}$$
If you try to plot this with Maple, you will get a result that isn't correct for the problem. It seems we lose information in going from (17) to (18).
What justifies writing (18) instead of (17)? If I recall correctly, I see (18) much more frequently than (17) in textbooks/notes/etc.
Numerical Example Parameters
Let's consider a concrete example.
Assume
$$m=1$$
$$k=1$$
$$d=1$$
$$\gamma=\frac{\omega_0}{4}$$
Then
$$A_0=\frac{kd}{m}=1$$
$$\omega_0=\sqrt{\frac{2k}{m}}=\sqrt{2}$$
$$\gamma=\frac{1}{4}$$
$$b=m\gamma=\frac{1}{4}$$
The steady-state solution is
$$x(t)=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}\sin{(\omega_d t-\beta)}$$
The amplitude of the steady-state solution is
$$B=\frac{A_0}{(\omega_0^2-\omega_d^2)^2-\gamma^2\omega_d^2}=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}$$
A plot of this function is
The phase lag of the steady-state solution is
$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)=arg\left (2-\omega_d^2+i\frac{\omega_d^2}{16}\right )$$
The amplitude maximum does not occur at ##\omega_0##. It occurs at
$$\omega_{d,max}=\sqrt{\frac{2\omega_0^2-\gamma^2}{2}}<\omega_0$$
My question is about the consideration of the initial conditions.
Suppose
$$\omega_d=\frac{\omega_0}{2}$$
Then, the steady-state solution is
$$x_p(t)=\frac{32}{73}\sin{(0.7t-0.34)}$$
The amplitude is ##32/73## and the phase lag is ##-0.34\text{rad}##.
I am interested in what is happening at ##t=0## here.
What is the meaning of the negative value of ##x_p(0)## here?
I guess that for the initial conditions implicit above, the homogeneous solution is just zero. But what does this mean exactly?
I think what I am trying to understand is the following.
First of all, is there more than one possible particular solution?
I know that the homogeneous solution has two parameters so there is an infinite number of those solutions.
Given a particular solution, and given initial conditions, we obtain the two parameters to pin down a homogeneous solution such that the general solution describes the motion of our mass with the initial conditions.
If we were to have started with a different particular solution, we would get other values for the homogeneous parameters (given the initial conditions).
Are there other particular solutions?
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