Two masses and spring oscillation

[Tanya Sharma]

OK...Please leave it for a while.

The potential consists of an elastic part and a Coulomb-part. Expand the potential function into Taylor series about the equilibrium point. Keep up to the second order term. From the equilibrium condition and from the coefficient of the second-order term, you get the equivalent force constant.

ehild

Could you show me how you would get the equivalent force constant using your above approach ?

 

[ehild]

The second derivative of the potential is the negative of the first derivative of force. It would not be a different approach.
Anyway, you have to choose a frame-of reference first.

Sorry, I have to leave now, but I am soon back. ehild

 

[Tanya Sharma]

Congratulations on your 10000 post

 

[ehild]

Thanks I did not notice. Back to equivalent force constant. The potential energy is a function of some variable x. Assume U(x) has a local extreme somewhere, at x₀. You want to find the behaviour of the system near x₀.
The Taylor series of a function f(x) about x₀ is defined as

T(x)=\sum_0^∞{f^{n}|_{x=x_0}\frac{(x-x_0)^n}{n!}}
where f⁽ⁿ⁾ means the n-th derivative of f(x), and you have to take all derivatives at x=x₀.

There are some criteria when you can do the expansion and the series converges to f(x).

Now we assume that U(x) can be expanded and we stop at the second-order term.

U(x)≈U(x_0)+\frac{df(x_0)}{dx} (x-x_0)+\frac{1}{2}\frac{d^2 f(x_0)}{dx^2}(x-x_0)^2

f(x) has local extrem at x₀, so the first derivative has to be zero. We can choose the zero of the potential energy at x₀. Our approximate potential function reduces to one term U(x)≈\frac{1}{2}\frac{d^2 f(x_0)}{dx^2}(x-x_0)^2

If the second derivative of the potential function is positive, it has a minimum at x₀. The potential is the same as that of a spring, and the effective spring constant is equal to the second derivative of U at the equilibrium point.

D=\frac{d^2 f(x_0)}{dx^2}

The force is negative gradient of the potential : F=-dU/dx = -D(x-x₀₎). You can switch over to the variable x-x₀=s, and the the equation of motion is mds²/dt²+Ds=0. The angular frequency of the SHM is ω=√(D/m) (in our case, use the reduced mass).

ehild

 

[Tanya Sharma]

I have a very basic question that when do we expand a function and keep terms up to first order term (i.e first two terms) just as in post #42 and when do we expand up to second order (i.e first three terms) just as in above post ?

 

[ehild]

You can ignore the second order term in an expansion if it and all the other terms are much smaller than the first order one. If the first order-term is zero, as in the case at the bottom of a potential well, you have to keep the second order term.

How many terms are needed, it depends on the deviation from x₀, the centre of the expansion.
In this problem, we were interested in small oscillation about an equilibrium position. |x-x₀|could be sufficiently small.

ehild

 

[Tanya Sharma]

You can ignore the second order term in an expansion if it and all the other terms are much smaller than the first order one. If the first order-term is zero, as in the case at the bottom of a potential well, you have to keep the second order term.

ehild

Okay...thanks for enlightening me :)

Another naive question

But what if the first order term is not present ?

What is approximate value of $(1-(\frac{d}{x})^2)^2$ under the assumption x<<d ? Should it be $1$ or should it be $1-2(\frac{d}{x})^2$ ?

 

[ehild]

Okay...thanks for enlightening me :)

Another naive question

But what if the first order term is not present ?

What is approximate value of $(1-(\frac{d}{x})^2)^2$ under the assumption x<<d ? Should it be $1$ or should it be $1-2(\frac{d}{x})^2$ ?

If the first term is not present expand to the second-order term.

Remember cosx. If x is small, usually it is approximated by 1 or by 1-x².

$(1-(\frac{d}{x})^2)^2$ under the assumption x<<d : if x<<d then d/x >> 1 and you can ignore 1.

If you mean (1-(x/d)²)² it is approximately 1-2(x/d)².

But it depends what else is in the expression you want to approximate. If it is 1-(1-(x/d)²)², you have to keep the second-order term. If it is 4+(1-(x/d)²)², for example, you might ignore (x/d)²

ehild

 

[Tanya Sharma]

I was initially hesitant (rather afraid) asking these questions .But after reading the explanations I am quite glad and very satisfied that I put my doubts in front of you .

Thanks ehild

 

[ehild]

I read just now : "the only stupid question is which is not asked".

ehild

 

[ehild]

Returning to your energy consideration, and assuming the CM is in rest: If you have two masses with coordinates x1 and x2 corresponding velocities v1 and v2, and you introduce the variable x=x2-x1, the corresponding velocity v=v2-v1:

CM is in rest: m1v1+m2v2=0, and v2-v1=v →
$v_1=-\frac{m_2v}{m_1+m_2}\\v_2=\frac{m_1v}{m_1+m_2}$

The kinetic energy is

$KE=\frac{1}{2} \left(m_1 (\frac{m_2 v}{m_1+m_2} )^2+ m_2( \frac{m_1 v}{m_1+m_2})^2 \right)$

$KE=\frac{1}{2}\frac{m_1 m_2}{m_1+m_2} v^2=\frac{1}{2} μ v^2$

Assuming SHM about the equilibrium value of x: x-xe=Asin(ωt), v=Aωsin(ωt) and expanding the potential energy about x_e: PE=0.5 D(x-xe)²

KE+PE=E →

$\frac{1}{2} A^2\left((μω)^2\cos^2(ωt)+D^2 \sin^2(ωt)\right)=E$

The expression can be constant only when μω=D.

 

[Tanya Sharma]

Hello ehild

Here is the derivation how a two-body problem can be reduced to one body-problem. I write it in one dimension,but the derivation is the same for 3D.

I would like to understand how this concept of reduced mass is applied in case of a Planet of mass m moving around sun of mass M in circular orbit

Suppose the Sun is considered stationary .In that case $\frac{GMm}{D^2}=\frac{mv^2}{D}$ where D is the distance between them .This gives $v=\sqrt{\frac{GM}{D}}$

Now if we want to take into account the motion of Sun also i.e both Sun and the planet orbiting about their common COM then this is equivalent to replacing the mass of planet 'm' by reduced mass 'μ = Mm/(M+m)' and replacing the Sun with a central mass 'M+m' .

Using this we have $\frac{G(M+m)\mu}{D^2}=\frac{mv^2}{D}$

Solving we get $v=\sqrt{\frac{GM}{D}}$ .

I am wondering why hasn't the result changed since now the Sun is also considered to be orbiting ?

 

[ehild]

What is v now?

ehild

 

[Tanya Sharma]

Speed of the planet in circular orbit.

 

[ehild]

No. The reduced-mass approach started by introducing new variables instead of the position vectors of the planet and Sun. You might look at my post in this thread about it.

ehild

 

[vela]

I would like to understand how this concept of reduced mass is applied in case of a Planet of mass m moving around sun of mass M in circular orbit

Suppose the Sun is considered stationary .In that case $\frac{GMm}{D^2}=\frac{mv^2}{D}$ where D is the distance between them .This gives $v=\sqrt{\frac{GM}{D}}$

By assuming the Sun is stationary, you've reduced the two-body problem to a one-body problem. In this case, you're assuming the Sun doesn't move because its mass M is so much larger than the mass of the planet.

Using the reduced-mass approach instead, you get
$$\frac{GMm}{D^2} = \frac{\mu v^2}{D}.$$ The force between the two bodies remains unchanged, but the mass of moving body is replaced by the reduced mass $\mu$. When $M \gg m$, you have $\mu \cong m$, and you recover the previous result.

Now if we want to take into account the motion of Sun also i.e both Sun and the planet orbiting about their common COM then this is equivalent to replacing the mass of planet 'm' by reduced mass 'μ = Mm/(M+m)' and replacing the Sun with a central mass 'M+m' .

I'm not sure how you came up with this.

 

[ehild]

The two-body Kepler problem can be solved by using the CM mass of reference. The difference of the position vectors obey the same differential equation as a single body with the reduced mass orbiting about a stationary mass of m+M. The gravitational force between the mass and the planet is F= GmM/D². That between the reduced mass and a central mass of m+M: G μ(M+m)/D²= G(mM/(m+M))(M+m)/D²=GmM/D².

ehild

 

[vela]

Ah, now I understand what Tanya was getting at in her post.

Tanya, you have a mistake in the equation of motion in the second approach, which is why you're ending up with identical results for $v$. You have to replace $m$ by $\mu$ on the righthand side.

 

[Tanya Sharma]

Ah, now I understand what Tanya was getting at in her post.

Tanya, you have a mistake in the equation of motion in the second approach, which is why you're ending up with identical results for $v$. You have to replace $m$ by $\mu$ on the righthand side.

$$ \frac{G(M+m)\mu}{D^2}=\frac{\mu v^2}{D} $$

$$ v^2 = \frac{G(M+m)}{D} $$

$$ v = \sqrt{\frac{GM(1+\frac{m}{M})}{D}} $$

Now ,since $m<<M$ , neglecting term $\frac{m}{M}$ , $1+\frac{m}{M} ≈ 1$

$$ v = \sqrt{\frac{GM}{D}} $$

But again this is the same result as before .

 

[vela]

Yes, that's correct. You see to that level of approximation, both approaches yield the same result, as you should expect. If you expand $\sqrt{1+\frac mM}$ as a series, you can calculate the higher-order corrections that come from relaxing the assumption that M is stationary.

 

[Tanya Sharma]

Suppose I do not consider this two body problem as a one body problem .

Let r₁ be the distance of M(sun) from CM and r₂ be distance of m(planet) from CM.Then $r_1 = \frac{mD}{m+M}$ and $r_2 = \frac{MD}{m+M}$

$$ \frac{GMm}{D^2} = \frac{mv^2}{r_2} $$

$$ \frac{GMm}{D^2} = \frac{m(m+M)v^2}{MD} $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

$$ v = \sqrt{\frac{GM}{(1+\frac{m}{M})D}} $$

Again $1+\frac{m}{M}≈ 1$

So,

$$ v = \sqrt{\frac{GM}{D}} $$

Is this analysis correct ?

 

[ehild]

The speed v what you get with the reduced mass approach is not the speed of the planet along its orbit about the CM. It is the speed of the hypothetical planet of mass μ orbiting the stationary Sun of mass m+M. Remember we introduced the variable r=r(planet)-r(Sun), and the time derivative of r is the relative velocity of the planet with respect to the Sun.

It is better to get the angular speed. It does not change with the reduced mass approach. So ω^2=G(M+m)/D³. The radii of the orbits both of the planet and the Sun is obtained from the condition that the CM is in rest in the applied frame of reference.


ehild

 

[vela]

$$ v = \sqrt{\frac{GM}{D}} $$

Is this analysis correct ?

Yes, but as ehild pointed out, this $v$ isn't the same $v$ as in your previous analyses.

 

[ehild]

Suppose I do not consider this two body problem as a one body problem .

Let r₁ be the distance of M(sun) from CM and r₂ be distance of m(planet) from CM.Then $r_1 = \frac{mD}{m+M}$ and $r_2 = \frac{MD}{m+M}$

$$ \frac{GMm}{D^2} = \frac{mv^2}{r_2} $$

$$ \frac{GMm}{D^2} = \frac{m(m+M)v^2}{MD} $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

$$ v = \sqrt{\frac{GM}{(1+\frac{m}{M})D}} $$

Again $1+\frac{m}{M}≈ 1$

So,

$$ v = \sqrt{\frac{GM}{D}} $$

Is this analysis correct ?

Yes, that is correct. But you want the difference between the cases with stationary Sun and with moving one. And what would you get for double stars, that is two nearly equal masses orbiting about their common CM? Like theese http://astrobob.areavoices.com/2010/04/10/how-to-double-your-enjoyment-of-the-heavens/
ehild

 

[Tanya Sharma]

But you want the difference between the cases with stationary Sun and with moving one.

Yes.

And what would you get for a double star, that is two nearly equal masses orbiting about their common CM?

$$ v = \sqrt{\frac{Gm}{2D}} $$ ,where m is the mass of a single star and D is the distance between them.

 

[Tanya Sharma]

Well.. it is quite confusing that both post#69 and post#71 are correct ,value of $v$ is coming out same ,yet the $v$ in the two posts are different .

Does $v$ in post#71 represent the linear speed of the planet in circular orbit around the CM ?

 

[ehild]

In post #69 you got the speed of the reduced-mass hypothetical planet:

v_r^2=G\frac {m+M}{D}
In post #71, you derived the speed of the real planet:

v^2=G\frac {M^2}{(M+m)D}

They are not the same.

 

[Tanya Sharma]

It is better to get the angular speed. It does not change with the reduced mass approach. So ω^2=G(M+m)/D³. The radii of the orbits both of the planet and the Sun is obtained from the condition that the CM is in rest in the applied frame of reference.

$$ \frac{G(M+m)\mu}{D^2}=\mu ω^2D $$

$$ ω^2 = \frac{G(M+m)}{D^3} $$

Now , $v = ωr_2$ where $r_2 = \frac{MD}{m+M}$

$$ v^2 = \frac{G(M+m)}{D^3} (\frac{MD}{m+M})^2 $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

This result matches the one in post#71 .

Is this how you wanted me to approach ?

 

[ehild]

$$ \frac{G(M+m)\mu}{D^2}=\mu ω^2D $$

$$ ω^2 = \frac{G(M+m)}{D^3} $$

Now , $v = ωr_2$ where $r_2 = \frac{MD}{m+M}$

$$ v^2 = \frac{G(M+m)}{D^3} (\frac{MD}{m+M})^2 $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

This result matches the one in post#71 .

Is this how you wanted me to approach ?

Yes, it was. That is the "real" speed of the planet. And the speed of the Sun centre is ωr₁

ehild

 

[Tanya Sharma]

Thank you very much :)

But I do not understand the importance of this reduced mass approach . We could have calculated the speed as in post#71 .

What could have been asked in the question where the conversion of two body problem in one would have been a superior approach ?

 

[ehild]

How would you solve the problem of double stars in general ( not assuming circular orbit)? When you need the period and the data of the orbits?

ehild

 

[Tanya Sharma]

OK.

Please have a look at the attachment .I have attached portion of a chapter in the book.

The motion of a planet is analogous to that of an electron revolving the nucleus just like in Bohr Model of the hydrogen atom .

If the nucleus is considered stationary then $$ \frac{1}{4\piε_0}\frac{e^2}{r^2}
= \frac{mv^2}{r} $$
But if take into account the motion of nucleus around the CM ,then do we simply replace $m$ by $\mu$ where $\mu =\frac{mM}{m+M}$ ,where m is the mass of electron and M is the mass of proton(nucleus).

The LHS in both the cases remain same whereas the RHS in the stationary case has $m$ and the moving nucleus case has $\mu$ .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{\mu v^2}{r} $$

But this is not correct because the 'v' in this case is that of the hypothetical reduced mass ,not the actual speed of the electron.

What is your opinion ? Is the book getting it wrong or am I interpreting it incorrectly ?

 

[ehild]

Here the force is proportional to the product of charges instead of the product of the masses.

ehild

 

[vela]

Well, it is quite confusing that both post#69 and post#71 are correct, value of $v$ is coming out same, yet the $v$ in the two posts are different .

Does $v$ in post#71 represent the linear speed of the planet in circular orbit around the CM ?

Try calculating $v_1$ and $v_2$, the speed of each body relative to the CM, using your approach in post 71. The bodies move in opposite directions, so their relative speed is given by $v_1+v_2$. Calculate that sum and compare it to the result you got in post 69. Don't make any approximations based on the assumption that $M \gg m$.

 

[Tanya Sharma]

Here the force is proportional to the product of charges instead of the product of the masses.

ehild

Right . But according to book ,if we need to account for the motion of nucleus we simply need to replace $m$ by $\mu$ to get the speed .(Or am I understanding it incorrectly)

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{\mu v^2}{r} $$

This gives $$v^2 = \frac{1}{4 \piε_0}\frac{e^2(m+M)}{rmM}$$

But this is incorrect value of speed .

The correct result can be obtained on similar lines to post#71 .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{mv^2}{r_2} $$ ,where $r_2 = \frac{Mr}{m+M}$ is the distance of the electron from the CM .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{mv^2}{r_2} $$

$$v^2 = \frac{1}{4 \piε_0}\frac{e^2M}{rm(m+M)}$$ . This is the correct speed .

How is book interpretation correct ?

 

[vela]

How is book interpretation correct ?

You're comparing apples to oranges. Reread the first part of post 72 and do the calculation I suggested in post 84. Hopefully, that will clear up your confusion.

 

[ehild]

The Bohr model predicts the spectrum of hydrogen atom. It postulates that the angular momentum is integer multiple of (h/2pi) and determines the possible energy levels. The angular momentum and the energy are those of the whole Hydrogen atom. The angular momentum and energy of the whole atom can be considered as those of a single particle with mass equal to the reduced mass of the electron-nucleus system. https://www.google.com/url?sa=t&rct...mG2jGkozqYJcq5A&bvm=bv.68445247,d.bGQ&cad=rjt

The reduced mass correction explains the isotope shift of the spectral lines of Deuterium. In case of stationary nuclei, the spectra of both the Hydrogen and the Deuterium atoms would be the same.

ehild