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Simple Harmonic equations.
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joyoustele
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Pangol
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joyoustele
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#3
(Original post by Pangol)
Are you aware of the connection between simple harmonic motion and circular motion? This is much easier to explain in terms of circular motion than straight SHM.
Are you aware of the connection between simple harmonic motion and circular motion? This is much easier to explain in terms of circular motion than straight SHM.
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Pangol
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Think about an object moving in a circle at a constant speed. It makes a certain number of full revolutions every second - this is the frequency, f, of the circular motion.
For every complete revolution, the object turns through an angle of 2 pi. So if it makes f complete revolutions per second, then it turns through an angle of 2 pi f radians per second. This is what we call angular frequency.
Now imagine a mass on a spring, bouncing up and down. You could imagine this as what it would look like if you looked at a point undergoing circular motion from the side on. There's a good animation of this at this Wikipedia URL, that I don't seem to be able to embed; https://en.wikipedia.org/wiki/Simple...tion_Orbit.gif
The angular frequency of the SHM is the angular frequency of this circular motion. You can also calculate it using 2 pi f, where f is the frequency of the SHM.
It's a bit odd for a book to do SHM before circular motion - the former follows so naturally from the latter!
For every complete revolution, the object turns through an angle of 2 pi. So if it makes f complete revolutions per second, then it turns through an angle of 2 pi f radians per second. This is what we call angular frequency.
Now imagine a mass on a spring, bouncing up and down. You could imagine this as what it would look like if you looked at a point undergoing circular motion from the side on. There's a good animation of this at this Wikipedia URL, that I don't seem to be able to embed; https://en.wikipedia.org/wiki/Simple...tion_Orbit.gif
The angular frequency of the SHM is the angular frequency of this circular motion. You can also calculate it using 2 pi f, where f is the frequency of the SHM.
It's a bit odd for a book to do SHM before circular motion - the former follows so naturally from the latter!
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joyoustele
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#5
(Original post by Pangol)
Are you aware of the connection between simple harmonic motion and circular motion? This is much easier to explain in terms of circular motion than straight SHM.
Are you aware of the connection between simple harmonic motion and circular motion? This is much easier to explain in terms of circular motion than straight SHM.
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K-Man_PhysCheM
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(Original post by joyoustele)
I have read up on circular motion.
I have read up on circular motion.
If we observe Hooke's Law
and note from Newton's Second Law that 
, we obtain the following second-order differential equation:
Where
Now this is a standard 2nd order DE, with the general solution:
We can use the boundary conditions (ie where the particle is at time
) to find the values of 
. Note that 
are amplitudes (and one will be zero) while 
are phasers that change the phase.Now
determines how long it will take to complete one full cycle. For example, if 
then the period of the each full cycle will be 2 seconds, since the period of the sinusoidal trig functions is 
... in fact, is the angular frequency!And since
, we see that the angular frequency increases as spring constant increases (so as spring gets stiffer, particle oscillates more quickly) and it decreases as mass increases (heavier particle takes longer for each oscillation than a lighter one on an identical same spring).We can combine this with the definition from the above poster and get that
, which may or may not be useful. I think if you are learning this as part of M3 or physics A-level, you don't need to know this much detail but it should hopefully help you understand where this all comes from
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joyoustele
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#7
(Original post by Pangol)
Think about an object moving in a circle at a constant speed. It makes a certain number of full revolutions every second - this is the frequency, f, of the circular motion.
For every complete revolution, the object turns through an angle of 2 pi. So if it makes f complete revolutions per second, then it turns through an angle of 2 pi f radians per second. This is what we call angular frequency.
Now imagine a mass on a spring, bouncing up and down. You could imagine this as what it would look like if you looked at a point undergoing circular motion from the side on. There's a good animation of this at this Wikipedia URL, that I don't seem to be able to embed; https://en.wikipedia.org/wiki/Simple...tion_Orbit.gif
The angular frequency of the SHM is the angular frequency of this circular motion. You can also calculate it using 2 pi f, where f is the frequency of the SHM.
It's a bit odd for a book to do SHM before circular motion - the former follows so naturally from the latter!
Think about an object moving in a circle at a constant speed. It makes a certain number of full revolutions every second - this is the frequency, f, of the circular motion.
For every complete revolution, the object turns through an angle of 2 pi. So if it makes f complete revolutions per second, then it turns through an angle of 2 pi f radians per second. This is what we call angular frequency.
Now imagine a mass on a spring, bouncing up and down. You could imagine this as what it would look like if you looked at a point undergoing circular motion from the side on. There's a good animation of this at this Wikipedia URL, that I don't seem to be able to embed; https://en.wikipedia.org/wiki/Simple...tion_Orbit.gif
The angular frequency of the SHM is the angular frequency of this circular motion. You can also calculate it using 2 pi f, where f is the frequency of the SHM.
It's a bit odd for a book to do SHM before circular motion - the former follows so naturally from the latter!
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joyoustele
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#8
(Original post by K-Man_PhysCheM)
I might just add a bit more to this discussion.
If we observe Hooke's Law and note from Newton's Second Law that , we obtain the following second-order differential equation:
Where
Now this is a standard 2nd order DE, with the general solution:
We can use the boundary conditions (ie where the particle is at time) to find the values of . Note that are amplitudes (and one will be zero) while are phasers that change the phase.
Now determines how long it will take to complete one full cycle. For example, if then the period of the each full cycle will be 2 seconds, since the period of the sinusoidal trig functions is ... in fact, is the angular frequency!
And since, we see that the angular frequency increases as spring constant increases (so as spring gets stiffer, particle oscillates more quickly) and it decreases as mass increases (heavier particle takes longer for each oscillation than a lighter one on an identical same spring).
We can combine this with the definition from the above poster and get that, which may or may not be useful.
I think if you are learning this as part of M3 or physics A-level, you don't need to know this much detail but it should hopefully help you understand where this all comes from
I might just add a bit more to this discussion.
If we observe Hooke's Law
and note from Newton's Second Law that , we obtain the following second-order differential equation:Where
Now this is a standard 2nd order DE, with the general solution:
We can use the boundary conditions (ie where the particle is at time
) to find the values of . Note that are amplitudes (and one will be zero) while are phasers that change the phase.Now
determines how long it will take to complete one full cycle. For example, if then the period of the each full cycle will be 2 seconds, since the period of the sinusoidal trig functions is ... in fact, is the angular frequency!And since
, we see that the angular frequency increases as spring constant increases (so as spring gets stiffer, particle oscillates more quickly) and it decreases as mass increases (heavier particle takes longer for each oscillation than a lighter one on an identical same spring).We can combine this with the definition from the above poster and get that
, which may or may not be useful. I think if you are learning this as part of M3 or physics A-level, you don't need to know this much detail but it should hopefully help you understand where this all comes from
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