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How can Simple Harmonic Motion have angular frequency? 
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#1
Feb1213, 09:45 AM

P: 79

It isn't making any intuitive sense. If it isn't moving in circular motion, how can it have angular frequency or speed?
Also, [tex]v=\pm ω\sqrt { A^{ 2 }x^{ 2 } }[/tex] only applies to SHM with springs only, right? Also, does anyone know how to derive this equation below? [tex]x=\frac { \pm \sqrt { { { v }_{ max } }^{ 2 }+{ v }^{ 2 } } }{ ω }[/tex] 


#2
Feb1213, 10:48 AM

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PF Gold
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The (differential) equation of motion for a particle of mass m on a spring with spring constant k and a displacement x from the equilibrium position is
m d_{2}x/dt^{2} = kx When you solve this, you find a solution of the form x= A sin(ωt  ∅) ∅ is an arbitrary value for the phase of the oscillation. That's where the variable ω comes from. Trig functions involve angles so ω has the dimension of an angle divided by time. Hence it's referred to as an angular frequency. There is one other point and that is that ω is in radians (as all good angles are) so is 2∏f, where f is the frequency in cycles per second (Hz). 


#3
Feb1213, 10:51 AM

P: 79

Also, since you solved the differential equation of motion for a particle with a spring, does that mean that the solution only applies to simple harmonic motion with springs? Is there simple harmonic motion without a spring? 


#4
Feb1213, 10:55 AM

P: 79

How can Simple Harmonic Motion have angular frequency?
http://www.youtube.com/watch?feature...h_Lr588#t=167s
He says that the tangential component of the force of gravity is mgsin(θ), but why is it negative? If the tangential component is in the same direction as the velocity, shouldn't it be positive? 


#5
Feb1213, 11:07 AM

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It doesn't have to be a spring. That equation of motion applies to many physical situations  where a 'restoring force' happens to be proportional to displacement from an equilibrium position. A simple pendulum approximates to this for small displacements from the vertical, too. Why not look it up????? BTW, the Negative sign in the equation is because the direction of the restoring force is opposite to the direction in which the particle has been displaced (i.e. back where it came from). If you are not familiar with equations of motion then I am sure you will come across them before too long and things may get clearer. 


#6
Feb1213, 11:13 AM

P: 79




#7
Feb1213, 11:24 AM

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When the position is the other side of the origin (i.e. the displacement is Negative), the force becomes positive. The situation is symmetrical for both sides, the force is always back to the equilibrium position. Draw it out for both cases.



#8
Feb1213, 11:30 AM

P: 79

I think I get why it's negative now.
By convention, we assign the right direction as the positive direction. Let's take a look at the term mgsin(θ). Assuming m and g are always positive, the term changes sign depending on what θ is. When θ is positive, sin(θ) is positive and the term mgsin(θ) is positive. But we know the restoring force is to the left when θ is positive, therefore, the force must be negative and so we add a negative sign. When θ is negative, sin(θ) is negative and the term mgsin(θ) becomes negative. But we know that the restoring force is to the right when θ is negative, therefore, the force must be positive, and so we add a negative sign. It works even when you assign the left as the positive direction. Part of why I didn't understand you was because I didn't actually understand displacement. Displacement is position away from the equilibrium. It is positive when the bob is to the right of the equilibrium and left when the bob is to the left of the equilibrium. For some reason, I thought displacement was instantaneous velocity. 


#9
Feb1213, 11:37 AM

P: 79

I've realized that a lot of simple harmonic motion is basically periodic motion in which the restoring force is proportional to the position away from the equilibrium. Does that mean all simple harmonic motion can be modelled using a spring?
The equation: [tex]ω=\sqrt { \frac { k }{ m } }[/tex] applies to simple harmonic motion when a spring is involved where k is the spring constant. But k doesn't need to be the spring constant right? It just needs to be the proportional constant of the restoring force to the displacement, right? [tex]k=\frac { F }{ x }[/tex] where F is any restoring force. Now my question is: is this ever useful when dealing with simple harmonic motion problem where the question doesn't involve a spring? The issue is that problems don't give you the proportional constant of the restoring force to the displacement when there is no spring involved I'm assuming? [tex]θ(t)={ θ }_{ max }cos(ωt+ϕ)[/tex] Does θ_{max} need to be in radians or degrees? It doesn't matter, does it? Will it matter if I take the derivative? [tex]w(t)={ θ }_{ max }ωsin(ωt+ϕ)[/tex] How can angular speed be a function of itself? 


#10
Feb1213, 11:42 AM

C. Spirit
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The general solution to the DE [itex]r'' + \omega ^{2}r = 0[/itex] would be [itex]r(t) = Ae^{i\omega t} + Be^{i\omega t} [/itex]. For simplicity of visualization, take A = 1, B = 0 so that [itex]r(t) = e^{i\omega t}[/itex]. You will recognize this as being [itex]exp:I\rightarrow S^{1}[/itex] where [itex]S^{1}[/itex] is the unit circle (we are viewing it as being a subset of the complex plane) and [itex]I[/itex] is an appropriate interval in [itex]\mathbb{R}[/itex]. Now, with regards to the physics of simple harmonic motion, we just take the real part of the above expression for [itex]r(t)[/itex] but you can visualize [itex]\omega [/itex], the angular frequency, as being the rate at which the unit circle is being swept out in the complex plane.



#11
Feb1213, 11:58 AM

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#12
Feb1213, 12:03 PM

P: 79

Thanks for the insight sophiecentaur and WannabeNewton.
FYI, the only questions left are: 


#13
Feb1213, 12:20 PM

C. Spirit
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Are you sure it is [itex]v_{max}^{2} + v^{2}[/itex] as opposed to [itex]v_{max}^{2}  v^{2}[/itex]? Otherwise you get that when [itex]v = v_{max}[/itex], [itex]\omega x = \pm \sqrt{2v_{max}^{2}}[/itex] but [itex]x(v = v_{max})[/itex] should be zero since the maximum speed will occur at the equilibrium position (I am assuming this is for simple harmonic motion and that [itex]x[/itex] is the displacement from equilibrium as usual).



#14
Feb1213, 12:23 PM

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#15
Feb1213, 12:28 PM

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#16
Feb1213, 12:30 PM

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I already told you that all angles should be in radians  or you keep getting 2pi all over the place. For a start  your derivative is only correct if ω is in radians. Look up differentiating trig functions  basics of. Degrees suck in Science. Radians will be used all over the Universe (literally) by any civilisation with Maths but degrees are totally arbitrary.
I don't know what those equations mean as you didn't define the variables so it is not possible to derive them. The "w" on the left hand side is not omega! And what is the theta? Is this describing a tortional vibration? Those angles aren't the same angles as the angular velocity in the vibration cycle. See how confusing it is when the variables aren't defined. 


#17
Feb1213, 12:36 PM

P: 79

[tex]θ(t)={ θ }_{ max }cos(ωt+ϕ)[/tex] 


#18
Feb1213, 01:01 PM

Mentor
P: 11,745

http://www.youtube.com/watch?v=SZ541Luq4nE 


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