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[Achintya]

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[PeroK]

Summary:: While coming across the general equation of SHM i wondered as to how the concept of Sin fn came into this equation...and how have we replaced k/m with (w^2). ?

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This was not hard to find:

https://en.wikipedia.org/wiki/Simple_harmonic_motion

 

[Achintya]

This was not hard to find:

https://en.wikipedia.org/wiki/Simple_harmonic_motion

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but sir my question is on what basis are we assuming this fact.

how would someone guess this...

 

[etotheipi]

What's the equation of motion for a block on the end of a horizontal spring?

 

[PeroK]

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but sir my question is on what basis are we assuming this fact.

how would someone guess this...

It's not a guess, it's just shorthand notation.

 

[Achintya]

It's not a guess, it's just shorthand notation.

can you be more clear...because i read many textbooks and i couldn't find the reason behind this assumption...we just assume it out of no where...

 

[Achintya]

What's the equation of motion for a block on the end of a horizontal spring?

Sir that is an example of SHM, so that will definitely follow the general equation of the SHM

 

[etotheipi]

can you be more clear...because i read many textbooks and i couldn't find the reason behind this assumption...we just assume it out of no where...

It's not assumed, it's derived. Write down the full $F_x = m\ddot{x}$ relation for the spring and see where it gets you!

 

[PeroK]

can you be more clear...because i read many textbooks and i couldn't find the reason behind this assumption...we just assume it out of no where...

It's not an assumption, it's a shorthand notation. Physics texts often write:
$$\omega \equiv \sqrt{\frac k m}$$
Where you can read $\equiv$ as "is defined to be".

 

[Achintya]

It's not assumed, it's derived. Write down the full $F_x = m\ddot{x}$ relation for the spring and see where it gets you!

sir i have seen the derivation but that does not contain the omega term unless and until we take this assumption

 

[etotheipi]

Ah okay, then @PeroK 's right.

In any case say you didn't switch to the shorthand notation, you'd still get $x=A\cos{(\sqrt{\frac{k}{m}} t + \phi)}$. It sort of makes sense that $\sqrt{\frac{k}{m}}$ has to be some sort of angular speed since that's what you get if you take the time derivative of the bit inside the $\cos$! You could try increasing $t + \Delta t$ and showing that to go one more period you need a certain value of $\Delta t$ which has to be the time period, and this makes things a bit more concrete.

 

[Achintya]

just check this out ...the solution of the differential equaltion came in the form of k/m...but it is only after the assumption that k/m=w^2 that we can see the w term in shm general equation

 

[PeroK]

just check this out ...the solution of the differential equaltion came in the form of k/m...but it is only after the assumption that k/m=w^2 that we can see the w term in shm general equation

It's not an assumption.

 

[Ibix]

Start with $\sin(\sqrt{k/m}\,t)$. At what time has the system advanced 1 radian? 2 radians? 3?

Now, if something has angular velocity $\omega$, at what time has it advanced 1 radian? 2 radians? 3?

So what's the angular velocity of that first system where $\sin(\sqrt{k/m}\,t)$?

 

[PeroK]

just check this out ...the solution of the differential equaltion came in the form of k/m...but it is only after the assumption that k/m=w^2 that we can see the w term in shm general equation

Let me try to answer this a different way. Suppose you were telling me about SHM and you wrote down your equation:
$$F = -kx$$
And I asked: "why are you assuming that the motion is in the x-direction?" What happens if the motion is not in the x-direction? Then your equation and everything you are doing is invalid.

Then, I might ask, "why are you assuming the spring constant is $k$"? What happens if the spring constant is $2k$? Then all your equations don't work.

How would you answer that?

 

[Achintya]

Let me try to answer this a different way. Suppose you were telling me about SHM and you wrote down your equation:
$$F = -kx$$
And I asked: "why are you assuming that the motion is in the x-direction?" What happens if the motion is not in the x-direction? Then your equation and everything you are doing is invalid.

Then, I might ask, "why are you assuming the spring constant is $k$"? What happens if the spring constant is $2k$? Then all your equations don't work.

How would you answer that?

but sir this is something we know that shm is a one dimensional motion, so i think it makes sense to write F=-kx..since it was observed experimentally that the force is varying linearly with displacement..

 

[PeroK]

but sir this is something we know that shm is a one dimensional motion, so i think it makes sense to write F=-kx..since it was observed experimentally that the force is varying linearly with displacement..

What happens if motion is in the $y$ direction? How do you know it's in the x-direction?

In other words, why are you using $x$?

 

[Achintya]

What happens if motion is in the $y$ direction? How do you know it's in the x-direction?

it hardly makes a difference even if the motion were in the y-direction...what we are sure about is that its a one dimensional motion...isn't it?

 

[PeroK]

it hardly makes a difference even if the motion were in the y-direction...what we are sure about is that its a one dimensional motion...isn't it?

Exactly, the motion could be in the y-direction. I want you to justify your assumption that motion is in the x-direction. Why use $x$ instead of $y$?

 

[Achintya]

Exactly, the motion could be in the y-direction. I want you to justify your assumption that motion is in the x-direction. Why use $x$ instead of $y$?

so what's the conclusion?

 

[PeroK]

so what's the conclusion?

The answer is that it's not an assumption, in the sense of something to assume without justification. The motion is assumed to be one-dimensional. It's not an additional assumption that the motion is in the x-direction. We take $x$ to be the direction of motion.

Likewise, we take the quantity $\omega$ to be the quantity $\sqrt{k/m}$. This is not an assumption.

It's so fundamental to mathematical physics that you can replace a complicated expression with a single letter that it's difficult to explain.

Anyway, if you don't like $\omega$, then whenever you see $\omega$ just replace it with $\sqrt{k/m}$. They are equivalent ways of writing the same thing.

 

[Achintya]

The answer is that it's not an assumption, in the sense of something to assume without justification. The motion is assumed to be one-dimensional. It's not an additional assumption that the motion is in the x-direction. We take $x$ to be the direction of motion.

Likewise, we take the quantity $\omega$ to be the quantity $\sqrt{k/m}$. This is not an assumption.

It's so fundamental to mathematical physics that you can replace a complicated expression with a single letter that it's difficult to explain.

Anyway, if you don't like $\omega$, then whenever you see $\omega$ just replace it with $\sqrt{k/m}$. They are equivalent ways of writing the same thing.

Sir actually the source of confusion here was that the omega already has some physical significance of its own...its not an arbitrary variable...so u see this gives a wrong impression

 

[PeroK]

Sir actually the source of confusion here was that the omega already has some physical significance of its own...its not an arbitrary variable...so u see this gives a wrong impression

That's a different question. It remains to be shown what physical significance $\omega$ or $\sqrt{k/m}$ has. For the moment they are just two ways of writing the same thing.

The wiki article, for example, goes on to prove that:
$$T = 2\pi \sqrt{m/k}$$
Hence, justifying the definition of $\omega$ in this case.

 

[Gaussian97]

Ok let's see, if you derive the equation of motion you get
$$\ddot{x}=-\frac{k}{m}x$$
wich has as general solution:
$$x(t)=A\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}$$
Ok until here? Let's find now the period $T$, that is defined as the smallest non-zero time that fulfils $x(t)=x(t+T)$ for ANY value of $t$ so, by definition
$$x(t+T)=A\sin{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} T + \phi_0\right)}=A\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}$$
$$\sin{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} T + \phi_0\right)}-\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}=0$$
Applying some trigonometric identities
$$2\sin{\left(\sqrt{\frac{k}{m}} \frac{T}{2}\right)}\cos{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} \frac{T}{2} + \phi_0\right)}=0$$
The only way this can be true for ANY $t$ is
$$\sin{\left(\sqrt{\frac{k}{m}} \frac{T}{2}\right)}=0 \Longrightarrow \sqrt{\frac{k}{m}} \frac{T}{2}=k\pi\Longrightarrow T = 2k\pi \sqrt{\frac{m}{k}}$$
Because we want the smalles non-zero time
$$T = 2\pi \sqrt{\frac{m}{k}}$$
This is the period, then the angular frequency is defined as
$$\omega = \frac{2\pi}{T}=\sqrt{\frac{k}{m}}$$

 

[Achintya]

Ok let's see, if you derive the equation of motion you get
$$\ddot{x}=-\frac{k}{m}x$$
wich has as general solution:
$$x(t)=A\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}$$
Ok until here? Let's find now the period $T$, that is defined as the smallest non-zero time that fulfils $x(t)=x(t+T)$ for ANY value of $t$ so, by definition
$$x(t+T)=A\sin{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} T + \phi_0\right)}=A\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}$$
$$\sin{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} T + \phi_0\right)}-\sin{\left(\sqrt{\frac{k}{m}} t + \phi_0\right)}=0$$
Applying some trigonometric identities
$$2\sin{\left(\sqrt{\frac{k}{m}} \frac{T}{2}\right)}\cos{\left(\sqrt{\frac{k}{m}} t+ \sqrt{\frac{k}{m}} \frac{T}{2} + \phi_0\right)}=0$$
The only way this can be true for ANY $t$ is
$$\sin{\left(\sqrt{\frac{k}{m}} \frac{T}{2}\right)}=0 \Longrightarrow \sqrt{\frac{k}{m}} \frac{T}{2}=k\pi\Longrightarrow T = 2k\pi \sqrt{\frac{m}{k}}$$
Because we want the smalles non-zero time
$$T = 2\pi \sqrt{\frac{m}{k}}$$
This is the period, then the angular frequency is defined as
$$\omega = \frac{2\pi}{T}=\sqrt{\frac{k}{m}}$$

Wow..nice .i understand it now

 

[kuruman]

Here is another way to look at this conceptually and without many equations. Imagine a turntable rotating with constant angular speed $\omega$ in the xy-plane. A peg is glued on the turntable at radius $R$ from its center. As you look down (along the z-axis) on the turntable, a set of equations that describes the motion of the peg is $x(t)=R \cos(\omega t)~;~~y(t)=R \sin(\omega t)$. However if you change your point of view and look at the turntable along the y-axis, you will not see any of the motion along the y-axis. You will only see the peg going back and forth along the x-axis according to $x(t)=R \cos(\omega t)$. Now imagine a horizontal spring-mass system with $k$ and $m$ chosen by you so that $\sqrt{k/m}=\omega$. If you set this spring-mass system into oscillation with amplitude $R$ so that at $t=0$ it is at $x=R$, you will see that the mass and the peg will execute completely synchronized motion.