• Support PF! Buy your school textbooks, materials and every day products Here!

Rearranging equation

  • #1

Homework Statement


The rate at which the mass–spring system loses energy to its surroundings is referred
to as the Q-value for the oscillator. The Q-value is defined as:

Q=2π(E/ΔE)

I need to find ΔE/E, which is the fractional energy loss per cycle of the oscillation.

How would I rearrange this so that ΔE/E = 2π/Q

Homework Equations


Q=2π(E/ΔE)

The Attempt at a Solution


I've no idea how to rearrange it so that E/ΔE is swapped round
 

Answers and Replies

  • #2
105
3
$$Q=2\pi(\frac{E}{\triangle E})$$
then
$$\frac{Q}{2\pi} =(\frac{E}{\triangle E})$$
$$\frac{2\pi}{Q}=(\frac{ \triangle E}{ E})$$
 
Last edited:
  • #3
$$Q=2\pi(\frac{E}{\triangle E})$$
then
$$\frac{Q}{2\pi} =(\frac{E}\{\triangle E})$$
$$\frac{2\pi}{Q}=(\frac{ \triangle E}\{ E})$$
Hi Maged, thanks for your reply. Apologies, this is all new to me. What do you mean when you say /frac?
 
  • #4
105
3
Hi Maged, thanks for your reply. Apologies, this is all new to me. What do you mean when you say /frac?
I have tried to write my answer in Latex preview
to make those staff clear for you ,,
But it seems there is something went wrong

any way ,,

I have written it again

$$Q=2\pi(\frac{E}{\triangle E})$$
$$\frac{Q}{2\pi}=(\frac{E}{\triangle E})$$
$$\frac{2\pi}{Q}=(\frac{\triangle E}{E})$$
 
  • #5
105
3
What it looks like now
is it understandable ?
(:
 
  • #6
105
3
I have fixed my problem though,
Reload the page ,
 
  • #7
What it looks like now
is it understandable ?
:)
Indeed it is, thanks very much. Appreciate your help.
 
  • #8
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
I have tried to write my answer in Latex preview
to make those staff clear for you ,,
But it seems there is something went wrong

any way ,,

I have written it again

$$Q=2\pi(\frac{E}{\triangle E})$$
$$\frac{Q}{2\pi}=(\frac{E}{\triangle E})$$
Now, take the reciprocal of each side ...
$$\frac{2\pi}{Q}=(\frac{\triangle E}{E})$$
 
  • #9
105
3
Now, take the reciprocal of each side ...
I have missed this point ,, it should be there ,,,

:)
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
Did you read the later postings in your thread 79065 ?

Point is that the 'definition' doesn't hold unless Q is very large.
Energy in an oscillator has to do with amplitude squared (e.g. with a spring: E = 1/2 k x2).

A damped oscillation is represented (see here ) by $$ z(t) = A e^{-\zeta\omega_0t}\; sin \left ( \sqrt{1-\zeta^2}\; \omega_0t + \phi \right )$$ and with

##\zeta \ll 1## so ##T \approx {2\pi\over \omega_0}## for ##\ \left( z(t+T)\right )^2 / \left( z(t)\right )^2\ ##

you get ##\ e^{-2 \zeta\; 2\pi} = e^{-2\pi / Q}\ ## if ##\ Q={1\over 2\zeta}##.

Again with Q large enough, this is ##1-{2\pi \over Q}\ .\ \ ## So ##{\Delta E\over E} = {2\pi \over Q}##


Now to your question: how to rearrange Q = 2π(E/ΔE) to get ΔE/E = 2π/Q :

multiply left and right by ΔE
divide left and right by E
divide left and right by Q

but is this really what you were asking :nb) ?
 
  • #11
I have missed this point ,, it should be there ,,,

:)
Thanks
Did you read the later postings in your thread 79065 ?

Point is that the 'definition' doesn't hold unless Q is very large.
Energy in an oscillator has to do with amplitude squared (e.g. with a spring: E = 1/2 k x2).

A damped oscillation is represented (see here ) by $$ z(t) = A e^{-\zeta\omega_0t}\; sin \left ( \sqrt{1-\zeta^2}\; \omega_0t + \phi \right )$$ and with

##\zeta \ll 1## so ##T \approx {2\pi\over \omega_0}## for ##\ \left( z(t+T)\right )^2 / \left( z(t)\right )^2\ ##

you get ##\ e^{-2 \zeta\; 2\pi} = e^{-2\pi / Q}\ ## if ##\ Q={1\over 2\zeta}##.

Again with Q large enough, this is ##1-{2\pi \over Q}\ .\ \ ## So ##{\Delta E\over E} = {2\pi \over Q}##


Now to your question: how to rearrange Q = 2π(E/ΔE) to get ΔE/E = 2π/Q :

multiply left and right by ΔE
divide left and right by E
divide left and right by Q

but is this really what you were asking :nb) ?
Indeed, it was what I was asking. With reference to the earlier post, it turns out the graph was difficult to read and I was advised to read it so that it fitted the equation.

Thanks for your reply though, appreciated,
 
  • #12
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
Did you read the later postings in your thread 79065 ?
Good catch! Ryan McDonald has yet to answer my question in my last post there. :(

Hopefully he will find time to do so shortly.

Point is that the 'definition' doesn't hold unless Q is very large.
That restriction wasn't clear to me in that thread. Discussion wasn't exhaustive, but I concluded that in a system showing stable oscillation, ΔE would represent the energy deficit the system must add during each cycle to maintain constant amplitude of oscillation, regardless of Q. Are you saying this would not be correct unless Q is "large"?
 
  • #13
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
@NascentOxygen: I've been looking at this thing as an undriven oscillator all the time. Mainly because of the way the OP in thread Calculating energy left in oscillator after 3 cycles was formulated.

However, in post #5 there it becomes clear that Ryan found his Q from the resonance curves, so what rude man states in his post #15 applies.

I must admit I have a hard time (=lazy me) coming up with a simple and elegant derivation for ΔE/E in the case of a driven damped oscillator.

@Ryan McDonald: 790685 post #13 should still be OK


@moderator: this whole thread should be appended to (or referred to in) thread 790685
 
  • #14
@NascentOxygen: I've been looking at this thing as an undriven oscillator all the time. Mainly because of the way the OP in thread Calculating energy left in oscillator after 3 cycles was formulated.

However, in post #5 there it becomes clear that Ryan found his Q from the resonance curves, so what rude man states in his post #15 applies.

I must admit I have a hard time (=lazy me) coming up with a simple and elegant derivation for ΔE/E in the case of a driven damped oscillator.

@Ryan McDonald: 790685 post #13 should still be OK


@moderator: this whole thread should be appended to (or referred to in) thread 790685
Thanks BvU. I found the question to be irritating and extremely ambiguous. Anyways, hopefully I've arrived at the correct solution now.
 
  • #15
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
If you share it we can look at it -- and perhaps comment :)

There's no reason to find the question irritating or ambiguous if there is a context.

And: any investment in this subject pays off for a lifetime in a vast number of disciplines, so cheer up!
 
  • #16
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
in post #5 there it becomes clear that Ryan found his Q from the resonance curves
I'm unsure whether fo/b.w. can be equated to Q for low Q's. I'd like to see that demonstrated as being true; meanwhile I'm regarding it as an approximation which holds for high Q's. After all, I expect it's only going to be exact for 2nd order systems, in any case.
 
  • #17
If you share it we can look at it -- and perhaps comment :)
Ok, here is my attempt.

I've taken Q to be 7 for plot a and 10 for plot b.

Rearranging Q=2π (E/ΔE)

ΔE/E = 2π/Q

For plot a:

After one cycle 2π/Q = 0.1

After two cycles (0.9*0.9) = 0.01

After three cycles = (0.81*0.9) = 0.001


For plot b:

After one cycle 2π/Q = 0.37

After two cycles = (0.63*0.63) = 0.23

After three cycles = (0.4*0.63) = 0.15

I'm sure there must be a more elegant way of calculating the energy left after 3 cycles but unfortunately, it evades me.

Thoughts welcome

Would it be 1-(2π/7)n for plot a? Which would show the amount of energy lost after n cycles
 
Last edited:
  • #18
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
Q to be 7 for plot a and 10 for plot b
Are these still from the picture 790685 post #5 ?
In that case I expect a to have a higher Q than b !!!

And what is ##\Delta\omega/\omega## ?

This one says "the bandwidth over which the power of vibration is greater than half the power at the resonant frequency, ωr = 2πfr is the angular resonant frequency, and Δω is the angular half-power bandwidth" but you are looking at an amplitude plot !

Then: your calculation: if q = 7, 2π/Q = 0.9, so after 1 cycle there is only 0.1 left. After 2 cycles 0.01 and after 3 cycles 0.001 !
 
  • #19
Are these still from the picture 790685 post #5 ?
In that case I expect a to have a higher Q than b !!!

And what is ##\Delta\omega/\omega## ?

This one says "the bandwidth over which the power of vibration is greater than half the power at the resonant frequency, ωr = 2πfr is the angular resonant frequency, and Δω is the angular half-power bandwidth" but you are looking at an amplitude plot !

Then: your calculation: if q = 7, 2π/Q = 0.9, so after 1 cycle there is only 0.1 left. After 2 cycles 0.01 and after 3 cycles 0.001 !
 
  • #20
Ah, silly me! I see my error now.

Yes they are from picture 790685 post #5

Q can be calculated from ω/Δω

where ω is the value at the peak of the curve

and Δω is the width of its peak at the halfway point

The graph is difficult to read but for plot b i originally took that to be 50/(55-45) which didn't work as q was less than 2π. I was advised to use a value of 7 to make the calculation work.
 
Last edited:
  • #21
If you share it we can look at it -- and perhaps comment :)
Are these still from the picture 790685 post #5 ?
In that case I expect a to have a higher Q than b !!!

And what is ##\Delta\omega/\omega## ?

This one says "the bandwidth over which the power of vibration is greater than half the power at the resonant frequency, ωr = 2πfr is the angular resonant frequency, and Δω is the angular half-power bandwidth" but you are looking at an amplitude plot !

Then: your calculation: if q = 7, 2π/Q = 0.9, so after 1 cycle there is only 0.1 left. After 2 cycles 0.01 and after 3 cycles 0.001 !
Sorry BvU, i've got myself a little confused! You are correct with regards to the values of the plots.

Plot a is 10 and plot b is 7.

It seems i'm a bit of a numpty today!
 
  • #22
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
Q can be calculated from ω/Δω
where ω is the value at the peak of the curve
Of the power curve. But your curve says amplitude at the vertical axis ! (is it amplitude ? What a coincidence both are same height !)
 
  • #23
Of the power curve. But your curve says amplitude at the vertical axis ! (is it amplitude ? What a coincidence both are same height !)
I'm having a nightmare with this question aren't I?!

We are advised that Q=(ω/Δω)

or Q=(f/Δf)

Can I work out Q using either of these equations from the information shown in the graph?

I'm so confused..
 
  • #24
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,014
You have a plot for the amplitude. The power is proportional to amplitude squared. So if the power is 1/2, the amplitude is 1/√2.

You want to take the width at maximum/√2
 
  • #25
You have a plot for the amplitude. The power is proportional to amplitude squared. So if the power is 1/2, the amplitude is 1/√2.

You want to take the width at maximum/√2
 

Related Threads on Rearranging equation

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
28
Views
42K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
Top