Two masses and spring oscillation

[ehild]

Expand f(x)=k_eq²/(x+L)² around x_e.

ehild

 

[Tanya Sharma]

OK..Taylor expanding f(x) around x_e and keeping terms upto first order

$$ f(x) = \frac{k_eq^2}{(x_e+L)^2} - \frac{2k_eq^2}{(x_e+L)^3}(x-x_e)$$

Now ,we had

$$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$

which can be rewritten as $$\mu \frac{d^2 x}{dt^2} = -kx + \frac{k_eq^2}{(x_e+L)^2} - \frac{2k_eq^2}{(x_e+L)^3}(x-x_e)$$

$$ \mu \frac{d^2 x}{dt^2} + (k+\frac{2k_eq^2}{(x_e+L)^3}) x = \frac{k_eq^2}{(x_e+L)^2} + \frac{2k_eq^2x_e}{(x_e+L)^3} $$

The effective spring constant is $$ K = k+\frac{2k_eq^2}{(x_e+L)^3} $$ .

Does this make sense ?

 

[vela]

We have two equations,

k_eq²/x_e² = k(x_e-L)

$$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$

Here x_e is equilibrium separation and x is the extension from the equilibrium length, but what is the relationship between the two?

I think it's better to define $x_e$ such that
$$\frac{k_e q^2}{(x_e+L)^2} = k x_e.$$ It's the value of $x$ such that the system is in equilibrium.

Now change variables from $x$ to $x' = x - x_e$. You should be able to show the differential equation becomes
$$\mu \frac{d^2 x'}{dt^2} = -k(x'+x_e) + \frac{k_e q^2}{(x'+x_e+L)^2}.$$ Now expand the second term to first order with the assumption that $x' \ll x_e+L$. After you simplify, you should find that the constant term drops out.

 

[ehild]

Use the equilibrium condition:

kx_e=k_e\frac{q^2}{(x_e+L)^2}

\mu \frac{d^2 x}{dt^2} = -kx + kx_e- \frac{2k_e q^2}{(x_e+L)^3}(x-x_e)=-k(x-x_e)- \frac{2k_e q^2}{(x_e+L)^3}(x-x_e)=-k(x-x_e)-k(x-x_e)\frac{2x_e}{x_e+L}
Write the equation in terms of u=x-x_e:

\mu \frac{d^2 u}{dt^2} =-ku\left(1+\frac{2x_e}{x_e+L}\right)

You know that xe>0 as the charges repel each other.

ehild

 

[Tanya Sharma]

I think I have confused myself by the usage of ‘x’ and x_e .Initially I was using ‘x’ as extension in the spring and x_e as equilibrium length of the spring .But as suggested by Vela in post#34 and by use of ehild in post#35 ,’x’ is being used as relative distance between the masses and x_e as extension in the spring from its natural length at equilibrium .

You can assume that x-xe is small with respect to xe+L.

The equilibrium length of the spring is x_e+L .The separation between masses at any instant is x ,so extension in the spring from equilibrium length is given by x-(x_e+L) .

So shouldn't the small oscillation condition be $x-x_e-L << x_e+L$ or $x-x_e << x_e+2L$ ?

 

[ehild]

I think I have confused myself by the usage of ‘x’ and x_e .Initially I was using ‘x’ as extension in the spring and x_e as equilibrium length of the spring .But as suggested by Vela in post#34 and by use of ehild in post#35 ,’x’ is being used as relative distance between the masses and x_e as extension in the spring from its natural length at equilibrium .

No, I used x as you did x=x2-x1-L, , and xe is its value at equilibrium.

The equilibrium length of the spring is x_e+L .The separation between masses at any instant is x ,

No, the separation between masses is x+L.

But it would be less confusing to use x=x2-x1 as variable.

ehild

 

[TSny]

The equilibrium length of the spring is x_e+L .The separation between masses at any instant is x ,so extension in the spring from equilibrium length is given by x-(x_e+L) .

Yes, this is confusing. You are taking x_e to be the amount of stretch of the spring from its natural length at the equilibrium position. But then you take x to be the separation between the masses. So, your x_e is not the equilibrium value of x.

If you take x to be the separation between the masses, then it would be natural to take x_e to be the separation between the masses in the equilibrium position.

With x defined as the separation between the masses, write the equation of motion in terms of x and L.

Then let s, say, represent displacement from equilibrium: s = x - x_e. (Here, x_e is the value of x at equilibrium.)

s is assumed "small". Rewrite the equation of motion in terms of s and approximate it to first order in s.

 

[Tanya Sharma]

Yes, this is confusing. You are taking x_e to be the amount of stretch of the spring from its natural length at the equilibrium position. But then you take x to be the separation between the masses. So, your x_e is not the equilibrium value of x.

If you take x to be the separation between the masses, then it would be natural to take x_e to be the separation between the masses in the equilibrium position.

With x defined as the separation between the masses, write the equation of motion in terms of x and L.

Then let s, say, represent displacement from equilibrium: s = x - x_e. (Here, x_e is the value of x at equilibrium.)

s is assumed "small". Rewrite the equation of motion in terms of s and approximate it to first order in s.

$$ \mu \frac{d^2 x}{dt^2} = -k(x-x_e) + k_e\frac{q^2}{x^2} $$

$$ \mu \frac{d^2 s}{dt^2} = -ks + k_e\frac{q^2}{(s+x_e)^2} $$

Equilibrium condition is given by $\frac{k_e q^2}{(x_e)^2} = k(x_e-L)$

Does it make sense ?

 

[TSny]

$$ \mu \frac{d^2 x}{dt^2} = -k(x-x_e) + k_e\frac{q^2}{x^2} $$

$$ \mu \frac{d^2 s}{dt^2} = -ks + k_e\frac{q^2}{(s+x_e)^2} $$

Your expression for the spring force in these two equations is incorrect. When x = x_e, there is a nonzero spring force balancing the electric force.

 

[ehild]

$$ \mu \frac{d^2 x}{dt^2} = -k(x-x_e) + k_e\frac{q^2}{x^2} $$

xe is not the same as the relaxed length of the spring. You equation should be $$ \mu \frac{d^2 x}{dt^2} = -k(x-L) + k_e\frac{q^2}{x^2} $$

Equilibrium means that distance between the masses when the net force is zero between them.
$$ 0 = -k(x_e-L) + k_e\frac{q^2}{x_e^2} $$

ehild

 

[ehild]

See attached graph. Fs is the spring force, Fe is the electric force, F is the resultant.

ehild

 

[Tanya Sharma]

$$ \mu \frac{d^2 x}{dt^2} = -k(x-L) + k_e\frac{q^2}{x^2} $$

$$ \mu \frac{d^2 s}{dt^2} = -k(s+x_e-L) + k_e\frac{q^2}{(s+x_e)^2} $$

Equilibrium condition is given by $\frac{k_e q^2}{(x_e)^2} = k(x_e-L)$

Now let $f(s) = k_e\frac{q^2}{(s+x_e)^2}$ .Taylor expanding the function about 0 ,we have f(s) = f(0)+f'(0)s

$$ f(0) = k_e\frac{q^2}{x_e^2} $$

$$ f'(0) = -2k_e\frac{q^2}{x_e^3}$$

$$ f(s) = k_e\frac{q^2}{(x_e)^2}-2k_e\frac{q^2}{(x_e)^3}s$$

$$ \mu \frac{d^2 s}{dt^2} = -ks -kx_e + kL+ k_e\frac{q^2}{(x_e)^2}-2k_e\frac{q^2}{(x_e)^3}s $$

$$ \mu \frac{d^2 s}{dt^2} + (k+2k_e\frac{q^2}{x_e^3})s = k_e\frac{q^2}{(x_e)^2} -kx_e+kL $$

Or, $$ \mu \frac{d^2 s}{dt^2} + (k+2k_e\frac{q^2}{x_e^3})s = 0 $$

Is it correct ?

Please say yes :shy:

 

[TSny]

Is it correct ?

Please say yes :shy:

Yes, that's correct!

Perhaps a bit shorter: The reduced-mass equation of motion has the form of a single particle of mass μ moving under a net force

F(x) = F_spring(x) + F_elec(x).

Let x_e be whatever value of x makes the net force zero.

Then, Taylor expand about x_e:

F(x) ≈ F(x_e) + F'(x_e)(x-x_e) = F'(x_e)(x-x_e) = F'(x_e) s

Thus, -F'(x_e) is the effective "spring constant".

[In ehild's graph, compare the slope of the blue curve at x = x_e with the slope of the black line.]

 

[Tanya Sharma]

Thank you so much ehild and TSny :) .

 

[Tanya Sharma]

How can we do this problem using energy approach i.e conservation of energy ?

$$ E(x) = \frac{1}{2}k(x-L)^2+\frac{k_eq^2}{x}+\frac{1}{2}2mv^2 $$

Taking time derivative of the above equation and equating $\frac{dE}{dt}=0$,we have

$$ k(x-L) \dot{x}- \frac{k_eq^2}{x^2} \dot{x}+2mv \dot{v} = 0 $$

Is it correct ? If yes ,how should I proceed ?

 

[ehild]

You have to specify what is v and what is x, and how they are related. It is about small oscillations, but an isolated system can move as a whole and v can be anything. The kinetic energy need not be twice the KE of one mass.

ehild

 

[Tanya Sharma]

You have to specify what is v and what is x, and how they are related.

You are right .

'x' is the relative separation of the masses and 'v' is the speed of the masses w.r.t origin .

Should I write individual EOM for the masses or is it possible to write energy equation in terms of reduced mass ?

The kinetic energy need not be twice the KE of one mass.
ehild

Doesn't symmetry require the two masses to have equal speeds and since the COM is at rest ,the KE is double of the KE of an individual mass ?

What is the right way to approach using energy method?

 

[ehild]

What symmetry do you speak about?
Where is the origin? In your first post you said it was to the left from m1. The CM is not in rest in arbitrary frames of reference.

If the origin is in the line connecting the masses, but arbitrary otherwise, and x=x2-x1, the KE is
KE=\frac{1}{2}(m_1x_1^2+m_2x_2^2)= \frac{1}{2}(m_1x_1^2+m_2(x+x_1)^2). In case m1=m2,
KE= \frac{1}{2}m(x_1^2+(x+x_1)^2)
What do you do with x1?





ehild

 

[Tanya Sharma]

$$ E(x) = \frac{1}{2}k(x_2 - x_1-L)^2+\frac{k_eq^2}{(x_2-x_1)}+\frac{1}{2}m_1\dot{x_1^2} +\frac{1}{2}m_2\dot{x_2^2} $$

Taking time derivative of the above equation and equating $\frac{dE}{dt}=0$,we have

$$ k(x_2-x_1-L) (\dot{x_2}-\dot {x_1})- \frac{k_eq^2}{(x_2-x_1)^2 } (\dot{x_2}-\dot {x_1})+m_1\dot{x_1}\ddot{x_1} + m_2\dot{x_2}\ddot{x_2} = 0 $$

Does this make sense ?

 

[ehild]

It is correct but of no use.
The energy is the first integral of the equation of motion. If you differentiate it you get back F=ma.
Choose a frame of reference first. If you assume the CM fixed, place the origin there. What does it mean to x1 and x2?

ehild

 

[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