joyoustele
Badges: 19
Rep:
?
#1
Report Thread starter 2 years ago
#1
Attachment 687216What is meant by the angular frequency?
0
reply
Pangol
Badges: 14
Rep:
?
#2
Report 2 years ago
#2
(Original post by joyoustele)
Attachment 687202What is meant by the angular frequency?
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.
0
reply
joyoustele
Badges: 19
Rep:
?
#3
Report Thread starter 2 years ago
#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.
Circular motion is in the next topic i need to cover. I will have a quick read now.
0
reply
Pangol
Badges: 14
Rep:
?
#4
Report 2 years ago
#4
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!
0
reply
joyoustele
Badges: 19
Rep:
?
#5
Report Thread starter 2 years ago
#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.
I have read up on circular motion.
0
reply
K-Man_PhysCheM
Badges: 15
Rep:
?
#6
Report 2 years ago
#6
(Original post by joyoustele)
I have read up on circular motion.
I might just add a bit more to this discussion.

If we observe Hooke's Law F = -kx and note from Newton's Second Law that F=ma = m\dfrac{\mathrm{d}^2x}{\mathrm{d  }t^2}, we obtain the following second-order differential equation:

m\dfrac{\mathrm{d}^2x}{\mathrm{d  }t^2} = -kx

\dfrac{\mathrm{d}^2x}{\mathrm{d}  t^2} = -\frac{k}{m}x

\dfrac{\mathrm{d}^2x}{\mathrm{d}  t^2} = - \omega ^2 x

Where \omega = \sqrt{\frac{k}{m}}

Now this is a standard 2nd order DE, with the general solution:

x = Asin(\omega t+\theta ) + Bcos(\omega t + \phi )

We can use the boundary conditions (ie where the particle is at time t=0) to find the values of A, B, \theta, \phi. Note that A,B are amplitudes (and one will be zero) while \theta, \phi are phasers that change the phase.

Now \omega determines how long it will take to complete one full cycle. For example, if \omega = \pi then the period of the each full cycle will be 2 seconds, since the period of the sinusoidal trig functions is 2\pi... in fact, \omega is the angular frequency!

And since \omega = \sqrt{\frac{k}{m}}, 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 2\pi f = \sqrt{\frac{k}{m}}, 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
0
reply
joyoustele
Badges: 19
Rep:
?
#7
Report Thread starter 2 years ago
#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!
Makes sense, I managed to find a video of the gif instead.Thanks
0
reply
joyoustele
Badges: 19
Rep:
?
#8
Report Thread starter 2 years ago
#8
(Original post by K-Man_PhysCheM)
I might just add a bit more to this discussion.

If we observe Hooke's Law F = -kx and note from Newton's Second Law that F=ma = m\dfrac{\mathrm{d}^2x}{\mathrm{d  }t^2}, we obtain the following second-order differential equation:

m\dfrac{\mathrm{d}^2x}{\mathrm{d  }t^2} = -kx

\dfrac{\mathrm{d}^2x}{\mathrm{d}  t^2} = -\frac{k}{m}x

\dfrac{\mathrm{d}^2x}{\mathrm{d}  t^2} = - \omega ^2 x

Where \omega = \sqrt{\frac{k}{m}}

Now this is a standard 2nd order DE, with the general solution:

x = Asin(\omega t+\theta ) + Bcos(\omega t + \phi )

We can use the boundary conditions (ie where the particle is at time t=0) to find the values of A, B, \theta, \phi. Note that A,B are amplitudes (and one will be zero) while \theta, \phi are phasers that change the phase.

Now \omega determines how long it will take to complete one full cycle. For example, if \omega = \pi then the period of the each full cycle will be 2 seconds, since the period of the sinusoidal trig functions is 2\pi... in fact, \omega is the angular frequency!

And since \omega = \sqrt{\frac{k}{m}}, 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 2\pi f = \sqrt{\frac{k}{m}}, 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
Thanks
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Hertfordshire
    All Subjects Undergraduate
    Sat, 22 Feb '20
  • Ravensbourne University London
    Undergraduate Open Day Undergraduate
    Sat, 22 Feb '20
  • Sheffield Hallam University
    Get into Teaching in South Yorkshire Undergraduate
    Wed, 26 Feb '20

People at uni: do initiations (like heavy drinking) put you off joining sports societies?

Yes (291)
65.54%
No (153)
34.46%

Watched Threads

View All