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In summary, the conversation discusses the differential equation of simple harmonic motion and the process of solving it. The concept of angular frequency is introduced and its relationship to the equation is explained. It is noted that textbooks often define angular frequency at the beginning for convenience, but it can also be derived from the equation. The conversation concludes with a mathematical explanation of how angular frequency is related to the properties of the sine function.
- #1
fireflies
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I was doing the differential equation of simple harmonic motion. At a time, to bring the equation, it simply said k/m=ω2
How does it come? Is there any proof?
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- #2
- #3
fireflies
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Well, that doesn't help. We get the differential equation puying omega square =k/m. Then we solve it. So, if we do it reverse we obviously will get omega square =k/m. But why do we put it in the first time?
- #4
EJC
- 42
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If
F=ma
and
F=−kxThen
ma=−kx
(by equating the forces.)
Which can be also written as
ma+kx=0or
a+kmx=0Now if x is displacement, differentiating once with respect to time will give you velocity of the spring and then differentiating again with respect to time will give acceleration.
Displacement of a spring can be given by
x=A∗Cos(ωt)where A is the Amplitude of motion and
ω
is the angular frequency
Now Differenting once will give velocity;
v=−AωSin(ωt)and again to give acceleration
a=−Aω2Cos(ωt)Now substituting our formula for Acceleration and displacement into our equation of motion
a+kmx=0Gives
−Aω2Cos(ωt)+kmACos(ωt)=0Which can be rearranged to;
A(−ω2+km)Cos(ωt)=0Can get rid of the
A
and
Cos(ωt)which leaves
−ω2+km=0which can be rearranged to
ω=km−−−√
- #5
fireflies
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EJC said:
a+kmx=0Now if x is displacement, differentiating once with respect to time will give you velocity of the spring and then differentiating again with respect to time will give acceleration.
Displacement of a spring can be given by
x=A∗Cos(ωt)where A is the Amplitude of motion and
ω
is the angular frequencyNow Differenting once will give velocity;
v=−AωSin(ωt)and again to give acceleration
a=−Aω2Cos(ωt)
It will be a + (k/m)x=0 and
x=Asin(omega*t), right?
Well, it makes sense in case of circular motion. But in case of mass spring, again, we put it before solving the later equation (x=A sin(omega*t)). So, the later comes. Why we put it here?
Same case for simple pendulum. We just put omega sqr = g/L and omega square = k/m and bring out equations. Why? Why it is so obvious that omega suare will be equal to these anyways?
- #6
nasu
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This is a notation or definition. You replace that ratio by a single parameter.
Later you will see that the meaning of this new parameter is the frequency of the motion.
When you first do the substitution, it is just a mathematical operation. You don't need to know that k/m is frequency squared. You just try to simplify the equation by replacing two parameters by just one.
If you don't want to do it, you don't need to. It is not a necessary step.
You can find the solutions in terms of k/m and when you see that the square root of this represent the angular frequency you may replace it by omega. Or not.
- #7
fireflies
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frequency means angular frequency right?
nasu said:
Later you will see that the meaning of this new parameter is the frequency of the motion.
Well I am trying that. How to find it out?
- #8
jtbell
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If you don't make the substitution ##\omega^2 = k/m## during the derivation, but keep on going anyway, you eventually end up with a solution that looks something like $$x = A \sin \left( \sqrt{\frac{k}{m}} t \right)$$ At that point you might say, "Aha, that ##\sqrt{k/m}## looks like an angular frequency", and at that point define ##\omega = \sqrt{k/m}##.
Most textbooks make that definition at the beginning because they're anticipating the answer that they're leading up to.
- #9
nasu
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What jtbell said.
- #10
EJC
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jtbell said:
If you don't make the substitution ##\omega^2 = k/m## during the derivation, but keep on going anyway, you eventually end up with a solution that looks something like $$x = A \sin \left( \sqrt{\frac{k}{m}} t \right)$$ At that point you might say, "Aha, that ##\sqrt{k/m}## looks like an angular frequency", and at that point define ##\omega = \sqrt{k/m}##.
Most textbooks make that definition at the beginning because they're anticipating the answer that they're leading up to.
Right. It becomes convenient in certain circumstances to just represent sqrt(k/m) as omega because in certain applications (i.e. SHO) it appears all the time, and it has units of angular frequency. It's the same concept as writing F instead of ma... they both have units of force, but one is notationally easier.
- #11
fireflies
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jtbell said:
If you don't make the substitution ##\omega^2 = k/m## during the derivation, but keep on going anyway, you eventually end up with a solution that looks something like $$x = A \sin \left( \sqrt{\frac{k}{m}} t \right)$$ At that point you might say, "Aha, that ##\sqrt{k/m}## looks like an angular frequency", and at that point define ##\omega = \sqrt{k/m}##.
That makes sense. Specially for circular motions. I know for circular motions it is angular freq. anyhow.
Is it obvious for other SHM too? (I know it's obvious since textbook says. But any other way too show that? Except the reason that unit will be the same?)
- #12
fireflies
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Maybe I am quite understanding the reason. All SHM and circular motion will give same kind of result, right?
- #13
nasu
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Once you have the solution of the equation, it is obvious that the square root is a frequency. You don't need even need to know what physical system is described by the equation. It is something related to the properties of the sin function.
If you have sin(t), the period is T=2π and the frequency is f=1/2π. This means that sin(t+T)=sin(t).
Now if you have
[tex] sin(\sqrt{\frac{k}{m}}t) [/tex] then the period is given by
[tex] T=2 \pi \sqrt{\frac{m}{k}} [/tex]
and the frequency is
[tex] f=\frac{1}{T} =\frac{1}{2 \pi} \sqrt{\frac{k}{m}} [/tex]
If you multiply the frequency by 2π you get the angular frequency
[tex] \omega=2 \pi f = \sqrt{\frac{k}{m}} [/tex]
You can see that i did not use any information about what kind of motion is described by that equation. Can be a mechanical SHO, the x component of a circular motion, or even some electrical signal.
- #14
fireflies
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Wow! Amazing derivation
FAQ: How Does Omega Squared Equal k/m in Simple Harmonic Motion?
How does omega square relate to k and m?
The relationship between omega square (ω²), the spring constant (k), and the mass (m) is given by the equation ω² = k/m. This equation is known as the angular frequency formula and it describes the frequency of oscillations of a mass on a spring.
What is the significance of omega square in relation to k and m?
The value of omega square (ω²) is directly proportional to the spring constant (k) and inversely proportional to the mass (m). This means that as the spring constant increases, the frequency of oscillations also increases, and as the mass increases, the frequency decreases. Omega square is an important parameter in understanding the behavior of a mass-spring system.
How is omega square derived from the equation ω² = k/m?
The equation ω² = k/m can be derived from the equation for the period of oscillations of a mass on a spring, which is T = 2π√(m/k). By squaring both sides and rearranging, we get ω² = k/m. This equation shows the relationship between the angular frequency (ω) and the spring constant (k) and mass (m).
Can the value of omega square change?
Yes, the value of omega square can change depending on the values of k and m. As mentioned before, omega square is directly proportional to k and inversely proportional to m. So, if the spring constant or the mass changes, the value of omega square will also change accordingly.
How does omega square affect the behavior of a mass-spring system?
The value of omega square determines the frequency of oscillations of a mass on a spring. A higher value of omega square means a higher frequency of oscillations, which results in faster and more energetic movements of the mass. On the other hand, a lower value of omega square leads to a lower frequency and slower movements of the mass. Omega square is a crucial factor in understanding the dynamics of a mass-spring system.
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