+- Southperry.net (https://www.southperry.net)
+-- Forum: Social (https://www.southperry.net/forumdisplay.php?fid=14)
+--- Forum: Rubik's Cube (https://www.southperry.net/forumdisplay.php?fid=58)
+--- Thread: A Mass Hanging from a Spring (/showthread.php?tid=40039)
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
Suppose that there is a mass hanging from a spring. Then the differential equation regarding this motion is:
m*x'' = -k*x + m*g
m*x'' + k*x = m*g
The convention is that downwards is positive and upwards is negative.
If we guess that x1 = e^(r*t) is a solution to the homogeneous equation:
m*x'' + k*x = 0
Then:
m*r^2*e^(r*t)+k*e^(r*t) = 0
e^(r*t)*(m*r^2+k)=0
r^2 = -k/m
r = +/- sqrt(-k/m)
r = +/- i*sqrt(k/m)
x1 = C1*e^(i*t*sqrt(k/m))+C2*e^(-i*t*sqrt(k/m))
x1 = (C1+C2)*cos(sqrt(k/m)*t)+(C1-C2)*i*sin(sqrt(k/m)*t)
C1=(C3-C4*i)/2, C2=(C3+C4*i)/2
x1 = C3*cos(sqrt(k/m)*t)+C4*i*sin(sqrt(k/m)*t)
If we guess that the solution to the particular equation:
m*x''+k*x=m*g
is in the form x2=A, where A is a constant:
k*A=m*g
A=m*g/k
x2=m*g/k
Then the general solution is:
x=C3*cos(sqrt(k/m)*t)+C4*sin(sqrt(k/m)*t)+m*g/k
Let's suppose that we pulled the spring some distance x(0) and released it. The initial condition is therefore:
x(0)=x0
x'(0)=0
x(0)=x0=C3-m*g/k
C3=x0+m*g/k
x'(t)=-sqrt(k/m)*C3*cos(sqrt(k/m)*t)+sqrt(k/m)*C4*cos(sqrt(k/m)*t)
x'(0)=0=sqrt(k/m)*C4
C4=0
Therefore, the complete solution to the initial value problem is:
x=(x0+m*g/k)*cos(sqrt(k/m)*t)+m*g/k
Mathematically, this works. Or at least, it should work out. However, logically, this doesn't make any sense. The mass is hanging in the presence of gravity, so the time it takes for the mass to fall should be shorter than the time it takes to rise. In other words, the mass should fall faster than it rises because it's in a gravitational field. Yet the equation implies that it doesn't. The first question is: wtp?
Also, a completely mathematical problem that I encountered is the conversion of this step:
x1 = C1*e^(i*t*sqrt(k/m))+C2*e^(-i*t*sqrt(k/m))
to sines and cosines. The first way is shown above, but I'll copy and paste it down here:
x1 = C1*e^(i*t*sqrt(k/m))+C2*e^(-i*t*sqrt(k/m))
x1 = (C1+C2)*cos(sqrt(k/m)*t)+(C1-C2)*i*sin(sqrt(k/m)*t)
C1=(C3-C4*i)/2, C2=(C3+C4*i)/2
x1 = C3*cos(sqrt(k/m)*t)+C4*i*sin(sqrt(k/m)*t)
The second way is:
x1 = C1*e^(i*t*sqrt(k/m))+C2*e^(-i*t*sqrt(k/m))
C1 = C5*^(C6*i), C2 = C7*e^(C8*i)
x1 = C5*e^(C6*i)*e^(i*t*sqrt(k/m))+C7*e^(C8*i)*e^(-i*t*sqrt(k/m))
x1 = C5*e^(i*t*sqrt(k/m)+i*C6)+C7*e^(-i*t*sqrt(k/m)+i*C8)
x1 = (C5+C7)*cos(sqrt(k/m)*t+C6)+(C5-C7)*i*sin(sqrt(k/m)*t+C8)
C3=(C9-C10*i)/2, C5=(C9+C10*i)/2
x1 = C9*cos(sqrt(k/m)*t+C6)+C10*sin(sqrt(k/m)*t+C7)
Both of them are mathematically correct, so hypothetically, they should also be mathematically equivalent. Yet I don't see how that's possible. Using the first simplification, the range is -(C3+C4)<x1<C3+C4. Using the second simplification, the range is -(C9+C10)< x1< C9+C10. |C9| and |C10| must be less than |C3| and |C4|, because a part of them went into shifting the phase of sine and cosine. The second question is: wtp?
tl;dr: What am I doing wrong?
A Mass Hanging from a Spring - Nalek - 2011-04-01
...April fools?
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
Screw April fools. This is a legit question that is bugging me.
A Mass Hanging from a Spring - Justin - 2011-04-01
I completely misread the thread title and thought you were saying something else. I go into the thread and find THIS. Bleh.
A Mass Hanging from a Spring - Imagine - 2011-04-01
....I would've used a better word choice on the title....
A Mass Hanging from a Spring - Nalek - 2011-04-01
Justin Wrote:I completely misread the thread title and thought you were saying something else. I go into the thread and find THIS. Bleh.
Imagine Wrote:....I would've used a better word choice on the title....
Oh I geddit now
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
What are you guys going on about? I don't see anything wrong with the title.
A Mass Hanging from a Spring - Imagine - 2011-04-01
2147483647 Wrote:What are you guys going on about? I don't see anything wrong with the title.
"A mass hanging".
....think about it.
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
A Mass Hanging from a Spring - Kalovale - 2011-04-01
2147483647 Wrote:What are you guys going on about? I don't see anything wrong with the title.
Spoiler
c
![[Image: church_mass_kneel.jpg]](http://www.stclementshrine.org/images/church_mass_kneel.jpg)
![[Image: water%20spring.jpg]](http://www.filterclean.co.uk/images/infospringimages/water%20spring.jpg)
A Mass Hanging from a Spring - Rick - 2011-04-01
It's clearly a stretch to interpret it as such, how about attempting to answer the question?
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
Okay. Wow. If you really think that I'm talking about people hanging themselves from some large pond, you're either grammatically inept or you're a complete idiot.
A Mass Hanging from a Spring
If the spring were a water spring, then it would be impossible for anyone or anything to hang from it. And if that were the case, then I would have titled this thread, "A Mass Drowning in a Spring", or "A Mass Suicide at a Spring". If I meant that a religious mass or a mob were hanging themselves in the vicinity of a spring, then I would have titled this "A Mass Hanging at a Spring", or "A Mass Hanging near a Spring". If I meant the seasonal Spring, then I would have titled it "A Mass Hanging in Spring", or "A Mass Hanging During Spring".
The only possible to interpret the statement is that a spring, which is a mechanical device that stores energy and oscillates when perturbed, has a mass hanging from it. And obviously, I'm not referring to a large group of people hanging themselves from a giant mechanical spring.
I often wonder how people pass middle school English without a basic understanding of prepositions and modifiers. I wonder how they even survive day to day life. It must be such a pain to locate give or receive instructions, which pretty much underlies everything.
A Mass Hanging from a Spring - Shidoshi - 2011-04-01
Think of a pendulus. Does it go faster when it's approaching its lower point than when going away from it? No.
You can get answers for the velocities of the mass by using conservation of energy to determine the speed at any given point.
Just consider the 3 types of energy involved: gravity potential, kinetic and spring potential.
A Mass Hanging from a Spring - 2147483647 - 2011-04-01
A pendulum swings the fastest at the bottom, since it gains the most energy at that point.
m*g*Δh = 1/2*m*v^2
If the starting height is defined at 0, the velocity when the pendulum reaches the bottom is:
v=sqrt(2*g*h)
Clearly, the largest velocity is at height h, or the bottom, because if y(t) is the height of the bottom of the pendulum at any time t, then 0<y(t)<h.
-----
Edit: I see what you're saying now. So that verifies that the form of my first solution is correct.
How about the form of the second question?
A Mass Hanging from a Spring - Kalovale - 2011-04-01
2147483647 Wrote:I thought a pendulum swings the fastest at the bottom, since it gains the most energy at that point.
I believe this is true. Potential energy converts fully into kinetic energy. (pardon terminologies, I didn't study it in English)
A Mass Hanging from a Spring - Hanabira.Kage - 2011-04-01
Kalovale Wrote:I believe this is true. Potential energy converts fully into kinetic energy. (pardon terminologies, I didn't study it in English)
This. Somewhat.
Assuming zero energy loss,
Reduction in Potential Energy = Gain in Kinetic Energy
So not all the Potential Energy gets converted. If the pendulum bob is still a certain height h above the ground at the bottom of its swing, it will still contain some Gravitational Potential Energy. Still, delta P.E. = delta K.E.
Simple as that.
Assuming zero energy loss,
Reduction in Potential Energy = Gain in Kinetic Energy
So not all the Potential Energy gets converted. If the pendulum bob is still a certain height h above the ground at the bottom of its swing, it will still contain some Gravitational Potential Energy. Still, delta P.E. = delta K.E.
Simple as that.
A Mass Hanging from a Spring - Shidoshi - 2011-04-01
Hanabira.Kage Wrote:This. Somewhat.
Assuming zero energy loss,
Reduction in Potential Energy = Gain in Kinetic Energy
So not all the Potential Energy gets converted. If the pendulum bob is still a certain height h above the ground at the bottom of its swing, it will still contain some Gravitational Potential Energy. Still, delta P.E. = delta K.E.
Simple as that.
mgh potential energy is based on a referential height. You could always say it's from the lowest part of the swing. If you want REAL potential energy then you should go with this one that accounts for how the gravitation field varies with your distance to the center of the earth:
V® = -GMm/(Rt + r) where r is your distance to surface of the Earth and Rt is the radius of the Earth.