How Does Omega Squared Equal k/m in Simple Harmonic Motion?

[fireflies]

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?

 

[EJC]

http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

 

[fireflies]

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?

 

[EJC]

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−−−√

 

[fireflies]

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)

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?

 

[nasu]

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.

 

[fireflies]

frequency means angular frequency right?

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?

 

[jtbell]

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.

 

[nasu]

What jtbell said.

 

[EJC]

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.

 

[fireflies]

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?)

 

[fireflies]

Maybe I am quite understanding the reason. All SHM and circular motion will give same kind of result, right?

 

[nasu]

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.

 

[fireflies]

Wow! Amazing derivation