Two masses and spring oscillation

  • #26
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Yes, and write the differential equation for x-xe.

ehild
You should have expand around the equilibrium.

ehild
We have two equations ,

keq2/xe2 = k(xe-L)

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

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

Should I substitute the value of L from first equation into the second .
 
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  • #27
ehild
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Do the Taylor-series expansion around xe. You can assume that x-xe is small with respect to xe+L.
Also note that $$\ddot x = d^2(x-x_e)/dt^2$$

ehild
 
  • #28
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Do the Taylor-series expansion around xe. You can assume that x-xe is small with respect to xe+L.
Also note that $$\ddot x = d^2(x-x_e)/dt^2$$

ehild
Sorry...I didn't understand this.I am not good with approximations and taylor expansions .Could you please elaborate on this ?
 
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  • #29
ehild
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Do you know what Taylor series is? f(x) = f(xe)+f'(xe)(x-xe)+0.5 f''(xe) (x-xe)2....Stop at the first-order term.

xe is constant. So the derivatives of x-xe are the same as the derivatives of x. You can write the differential equation in terms of Δx=x-xe.

ehild
 
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  • #30
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Do you know what Taylor series is? f(x) = f(xe)+f'(xe)(x-xe)+0.5 f''(xe) (x-xe)2....Stop at the first-order term.
OK...I know the definition but haven't applied it in the problems.

What is f(x) which we need to expand ?
 
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  • #31
ehild
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Expand f(x)=keq2/(x+L)2 around xe.

ehild
 
  • #32
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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 ?
 
  • #33
vela
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We have two equations,

keq2/xe2 = k(xe-L)

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

Here xe 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.
 
  • #34
ehild
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Use the equilibrium condition:

[tex]kx_e=k_e\frac{q^2}{(x_e+L)^2}[/tex]

[tex] \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}[/tex]
Write the equation in terms of u=x-xe:

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

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

ehild
 
  • #35
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I think I have confused myself by the usage of ‘x’ and xe .Initially I was using ‘x’ as extension in the spring and xe 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 xe 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 xe+L .The separation between masses at any instant is x ,so extension in the spring from equilibrium length is given by x-(xe+L) .

So shouldn't the small oscillation condition be ##x-x_e-L << x_e+L ## or ##x-x_e << x_e+2L ## ?
 
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  • #36
ehild
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I think I have confused myself by the usage of ‘x’ and xe .Initially I was using ‘x’ as extension in the spring and xe 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 xe 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 xe+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
 
  • #37
TSny
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The equilibrium length of the spring is xe+L .The separation between masses at any instant is x ,so extension in the spring from equilibrium length is given by x-(xe+L) .
Yes, this is confusing. You are taking xe 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 xe 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 xe 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 - xe. (Here, xe 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.
 
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  • #38
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Yes, this is confusing. You are taking xe 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 xe 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 xe 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 - xe. (Here, xe 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 ?
 
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  • #39
TSny
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$$ \mu \frac{d^2 x}{dt^2} = -k(x-x_e) + k_e\frac{q^2}{x^2} $$

$$ \mu \frac{d^2 s}{dt^2} = -ks + k_e\frac{q^2}{(s+x_e)^2} $$
Your expression for the spring force in these two equations is incorrect. When x = xe, there is a nonzero spring force balancing the electric force.
 
  • #40
ehild
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$$ \mu \frac{d^2 x}{dt^2} = -k(x-x_e) + k_e\frac{q^2}{x^2} $$
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
 
  • #41
ehild
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See attached graph. Fs is the spring force, Fe is the electric force, F is the resultant.

ehild
 

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  • #42
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$$ \mu \frac{d^2 x}{dt^2} = -k(x-L) + k_e\frac{q^2}{x^2} $$

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

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:
 
  • #43
TSny
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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) = Fspring(x) + Felec(x).

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

Then, Taylor expand about xe:

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

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

[In ehild's graph, compare the slope of the blue curve at x = xe with the slope of the black line.]
 
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  • #44
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Thank you so much ehild and TSny :) .
 
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  • #45
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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 ?
 
  • #46
ehild
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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
 
  • #47
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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?
 
  • #48
ehild
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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
[tex]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)[/tex]. In case m1=m2,
[tex]KE= \frac{1}{2}m(x_1^2+(x+x_1)^2)[/tex]
What do you do with x1?????





ehild
 
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  • #49
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$$ E(x) = \frac{1}{2}k(x_2 - x_1-L)^2+\frac{k_eq^2}{(x_2-x_1)}+\frac{1}{2}m_1\dot{x_1^2} +\frac{1}{2}m_2\dot{x_2^2} $$

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 ?
 
  • #50
ehild
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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
 

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