Oscillators and conservation of energy

In summary, the author is explaining how a spring oscillator works, and he is using equations 7.4 and 7.5. equation 7.4 is about how the oscillator's energy is determined, and equation 7.5 is about how the oscillator's amplitude changes over time.
  • #1
velvetmist
15
1
In the equation 7.4, the author is taking v0=√(C/M)*x, and I don't get where does that come from. I would really appreciatte your help, thanks.
 

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  • #2
Could you please define the terms in the equation you posted? This is for a spring oscillator, right? Thanks.

Can you say what you think is being expressed with this equation? What is the total energy of a simple spring oscillator made up of?
 
  • #3
velvetmist said:
In the equation 7.4, the author is taking v0=√(C/M)*x
And you posted equation 7.5.
 
  • #4
berkeman said:
And you posted equation 7.5.
Sorry, I made a typo, I posted the correct one.

berkeman said:
What is the total energy of a simple spring oscillator made up of?
Yes.

berkeman said:
This is for a spring oscillator, right?
Yes.

berkeman said:
Could you please define the terms in the equation you posted?
It doesn't say much, cause it's trying to explain spring oscillators in a general way.
 

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  • #5
velvetmist said:
v0=√(C/M)*x

Maybe I missed it, but I don't see this in any of the stuff you posted.
 
  • #6
George Jones said:
Maybe I missed it, but I don't see this in any of the stuff you posted.
If you do the integral you got:
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. Then -M*v02=C*x2. So v0=√(-C/M)*x.

Edit: i just realized i forgot to put the minus.
 

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Last edited:
  • #7
velvetmist said:
If you do the integral you got: View attachment 231123. Then -M*v02=C*x2. So v0=√(-C/M)*x.

Edit: i just realized i forgot to put the minus.
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/

For example, $$\frac 1 2 Mv^2 - \frac 1 2 Mv_0^2$$
 
  • #8
jbriggs444 said:
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/

For example, $$\frac 1 2 Mv^2 - \frac 1 2 Mv_0^2$$

velvetmist said:
If you do the integral you got:
214204-13e95b0f6e34e0b7da0f6a7bf3dd9a7d.jpg
. Then -M*v02=C*x2. So v0=√(-C/M)*x.

$$ -M v^2_0 = C x^2 .$$ Therefore $$ v_0 = \sqrt {\frac {-C} {M}} x .$$

Btw, the text defines ##w_0 = \sqrt {\frac {C} {M}}, v_0 = w_0 A \cos(\phi). ## Then if ##v_0 = \sqrt {\frac {-C} {M}} x, x = A \cos(\phi). ## But we got that ## x = A \sin(\phi),## so ##\cos(\phi)## would be equal to ##\sin(\phi)## and that's not true. And even if it were true, I'm not taking into account that ##w_0 = \sqrt {\frac {C} {M}},## and not equal to ##\sqrt {\frac {-C} {M}}.##
 

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  • #9
Okay, now that I have figured out that I have a copy of the text (used in the first mechanics course I took all those decades ago), maybe we can have a discussion.

I don't understand what you mean when you write

velvetmist said:
If you do the integral you got ##\frac{1}{2} Mv^2 - \frac{1}{2} Mv_0^2 =##.

Do what integral? Also, this only half of an equation!

velvetmist said:
Then ##-M v^2_0 = C x^2##

I don't see how you got this.

Equation (7.5) in the text is
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2 = E,$$
where ##E## is the constant ##\frac{1}{2} CA^2##. Here, ##C## is the spring constant and ##A## is the amplitude of oscillation..

This means that
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} CA^2$$
at all times ##t##. Setting ##t=0## gives
$$\frac{1}{2} Mv_0^2 + \frac{1}{2} Cx_0^2 = \frac{1}{2} CA^2,$$
so that
$$v_0^2 = \frac{C}{M} \left( A^2 - x_0^2 \right) = \omega_0^2 \left( A^2 - x_0^2 \right).$$

Also, I do not understand

velvetmist said:
But we got that ## x = A \sin(\phi)##

Actually, ##x = A \sin \left( \omega_0 t + \phi \right)##.
 
  • #10
George Jones said:
Do what integral? Also, this only half of an equation!
(5.12) integral. And yes, i just realized that I cropted it in a wrong way, but basically: $$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2.$$

George Jones said:
I don't see how you got this.
Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$
and that's only true if ##−Mv_0^2=Cx^2.##

George Jones said:
Actually, ##x = A \sin \left( \omega_0 t + \phi \right)##.
I'm just taking that from the textbook. I already posted that part but i will do it again if that helps u:
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  • #11
velvetmist said:
Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$

I agree with
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 ,$$
but I do not understand the second equals sign, I do not understand why
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 $$
is also equal to
$$\frac 1 2 Mv^2-\frac 1 2 Cx^2 .$$

velvetmist said:
I'm just taking that from the textbook.

No, you have not taken it from the textbook. You wrote (and used)

velvetmist said:
But we got that ## x = A \sin(\phi)##

which is incorrect, and which is not in the textbook.
 
  • #12
George Jones said:
I agree with
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 ,$$
but I do not understand the second equals sign, I do not understand why
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2$$
is also equal to
$$\frac 1 2 Mv^2-\frac 1 2 Cx^2 .$$
Sorry, sorry, I'm really sorry, i just made a lot of typos, I meant
$$\frac 1 2 Mv^2+\frac 1 2 Cx^2 .$$
That's because
$$ \Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here ##\Delta V=0.## And we also know that
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2 = E,$$
then
$$\frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2.$$

George Jones said:
No, you have not taken it from the textbook. You wrote (and used)
Sorry, i just didn't read "at t=0", which is a huge mistake, but even like that i think is kind of weird cause he have that
$$ v_0 = \sqrt {\frac {C} M} A \cos(\phi) = \sqrt {\frac {-C} M}x .$$
And we know that
$$x = A \sin \left( \omega_0 t + \phi \right).$$
Then
$$v_0 = \sqrt {\frac {C} M} A \cos(\phi) = \sqrt {\frac {-C} M} A \sin \left( \omega_0 t + \phi \right).$$
$$\sqrt {-1}\cos(\phi)=\sin \left( \omega_0 t + \phi \right).$$
I probably made a mistake, or something like that cause i find this equation really weird.
 
  • #13
velvetmist said:
That's because
$$ \Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here ##\Delta V=0.##

No,
$$\Delta V = \frac{1}{2} C x^2 - \frac{1}{2} C x_0^2 .$$

See equation (5.20).
 
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  • #14
George Jones said:
No,
$$\Delta V = \frac{1}{2} C x^2 - \frac{1}{2} C x_0^2 .$$

See equation (5.20).
Thank you so much! I finally get it, all of my questions were between pages 149 and 150. We will see conservation of energy in three classes, that's why I was so confused about that.
 

1. What is an oscillator?

An oscillator is a physical system that exhibits periodic motion, meaning it repeats the same pattern in regular intervals. This can be seen in various systems such as a pendulum, a spring, or an electronic circuit.

2. How does an oscillator conserve energy?

Oscillators conserve energy through the principle of conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another. In an oscillator, energy is constantly being transformed between kinetic energy (motion) and potential energy (stored energy) as it oscillates back and forth.

3. What factors affect the period of an oscillator?

The period of an oscillator, or the time it takes to complete one full cycle, is affected by several factors including the mass, stiffness, and length of the oscillator. In general, a heavier mass or a stiffer oscillator will have a longer period, while a longer oscillator will have a shorter period.

4. Can an oscillator ever stop oscillating?

In a perfect system, an oscillator would continue to oscillate forever without losing any energy. However, in real-world systems, there is always some friction or resistance present that will eventually cause the oscillator to lose energy and stop oscillating.

5. How are oscillators used in everyday life?

Oscillators have many practical uses in everyday life, such as in clocks, watches, and other timekeeping devices. They are also used in musical instruments, radio and television transmitters, and many other electronic devices. Additionally, the concept of oscillation is important in fields such as engineering, physics, and biology.

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