Two masses and spring oscillation

  • #1
1,540
134

Homework Statement



The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is

(a) ≥ 1
(b) > 1
(c) ≤ 1
(d) = 1


Homework Equations





The Attempt at a Solution



First I calculated the time period of neutral masses .

Let us consider origin to be at the left of m1 .Take the x axis to be along the length of the spring and call the left mass be m1 at coordinate x1 and right mass be m2 at coordinate x2 .

x=x2-x1-L

d2x/dt2 = d2(x2-x1)/dt2

EOM for mass m1 is m1d2x1/dt2 = kx

EOM for mass m2 is m2d2x2/dt2 = -kx

After manipulating the above two equations we arrive at d2/dt2 + (k/μ)x = 0 ,where μ = m1m2/(m1+m2) .

The time period is 2π√(μ/k) .

Now when the masses are charged ,

EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2

EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

Is this the correct way to approach the problem ? If yes ,how should I proceed .
 

Attachments

Last edited:

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
This is just one of a few dozen multi-choice questions on an exam paper? If that's the case, there is probably not a lot of calculation to be done.

You are not told whether in this test the weights are changed from a smaller mass to a larger, or whether it's vice-versa? So even if the period were observed to change, how could you say whether the corresponding ratio is >1 when it might be <1 ?

So, purely on the absence of information allowing you to decide on any contrary option, it seems you are confined to only one answer.

insulated spring and mass system before and after charging
spelling: probably "isolated" spring, and after "changing" :smile:
 
Last edited:
  • #3
ehild
Homework Helper
15,523
1,901
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
 
  • #4
1,540
134
Have I done it correctly in the OP ? Have I correctly written the EOM for the two charged masses ?Could you help me proceed using the force method .
 
  • #5
1,948
200
This is just one of a few dozen multi-choice questions on an exam paper? If that's the case, there is probably not a lot of calculation to be done.

You are not told whether in this test the weights are changed from a smaller mass to a larger, or whether it's vice-versa? So even if the period were observed to change, how could you say whether the corresponding ratio is >1 when it might be <1 ?

So, purely on the absence of information allowing you to decide on any contrary option, it seems you are confined to only one answer.


spelling: probably "isolated" spring, and after "changing" :smile:
No, they really meant insulated and charging. Look at the picture. The question is whether charging the masses with electric charges q affects the osculation period.
 
  • #6
ehild
Homework Helper
15,523
1,901
Have I done it correctly in the OP ? Have I correctly written the EOM for the two charged masses ?Could you help me proceed using the force method .
It looks correct. Expand the force about the equilibrium point.

ehild
 
  • #7
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
No, they really meant insulated and charging. Look at the picture. The question is whether charging the masses with electric charges q affects the osculation period.
Big oops! I guess I should click on the thumbnail. :redface:
 
  • #8
1,540
134
It looks correct. Expand the force about the equilibrium point.
I am not sure what do we mean by equilibrium point in this two body problem ?

At equilibrium keq2/(x2e-x1e)2 = k(x2e-x1e-L) ,where x2eand x1e are respective equilibrium positions .

Or, keq2/xe2 = k(xe-L) ,where xe is the separation between the masses at equilibrium.

Is it correct ?


EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2

EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

Do I need to expand the 2nd term on the RHS about xe ?
 
Last edited:
  • #9
ehild
Homework Helper
15,523
1,901
I am not sure what do we mean by equilibrium point in this two body problem ?

At equilibrium keq2/(x2e-x1e)2 = k(x2e-x1e-L) ,where x2eand x1e are respective equilibrium positions .

Or, keq2/xe2 = k(xe-L) ,where xe is the separation between the masses at equilibrium.

Is it correct ?


EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2

EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

Do I need to expand the 2nd term on the RHS about xe ?
Yes, and write the differential equation for x-xe.

ehild
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,758
1,354
Do you know about reduced mass and converting a two-body problem into a one-body problem?
 
  • #11
1,540
134
Hello Vela

Do you know about reduced mass and converting a two-body problem into a one-body problem?
Please check my understanding about reduced mass .

In the two body problem ,just as in case 1 where uncharged masses are present and the two masses are under the action of internal spring force ,we may consider mass 1 to be stationary (or treat it as origin) and replace mass 2 with reduced mass μ = m1m2/(m1+m2) .The relative separation 'x' works as the x coordinate of the 2nd mass .

Is this correct ?

I initially intended using the concept of reduced mass ,but there may be something wrong with my implementation . Please have a look at the OP where I have found the time using reduced mass .

What is the correct way of converting this two body problem in one body problem ?
 
  • #12
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,758
1,354
Yes, that's correct, and the differential equation you derived reflects just what you said. The point is, you didn't need to rederive this result. You could have gone straight to
$$\mu \frac{d^2 x}{dt^2} = -kx.$$ For the charged masses, tack on the electric force and expand to first order to find the effective spring constant.
 
  • #13
1,540
134
You could have gone straight to
$$\mu \frac{d^2 x}{dt^2} = -kx.$$ For the charged masses, tack on the electric force and expand to first order to find the effective spring constant.
$$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$

Is this correct ?
 
  • #14
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,758
1,354
Looks good.
 
  • #15
1,540
134
When x<<L , $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{L^2}(1-\frac{2x}{L})$$

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

Here effective spring constant ##K = k+\frac{2k_eq^2}{L^3}## and time period is ## T = 2\pi \sqrt{\frac{\mu}{K}}##

Is this correct ?

Edit :Fixed errors
 
Last edited:
  • #16
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,758
1,354
Right idea, but you made some algebra mistakes.
 
  • #17
1,540
134
OK..I have fixed the errors .Does post#15 make sense ?
 
  • #18
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,758
1,354
Yup.
 
  • #19
1,540
134
OK...

Can I conclude that the time period of oscillation of charged masses is less than that of the uncharged masses i.e the correct option is b) >1 ?
 
Last edited:
  • #20
ehild
Homework Helper
15,523
1,901
Yes.

ehild
 
  • #21
1,540
134
Thanks...

Is the time period of md2x/dt2 + kx = 0 and md2x/dt2 + kx = C same ? Is the time period in both the cases given by 2π√(m/k) ?
 
  • #22
TSny
Homework Helper
Gold Member
12,765
3,124
Is the time period of md2x/dt2 + kx = 0 and md2x/dt2 + kx = C same ? Is the time period in both the cases given by 2π√(m/k) ?
Yes.

---------------------------

Maybe I'm just being a nuisance, but in your equation $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$ it is not necessarily true that x<< L for small oscillations.

If the spring constant is weak while the charge q is strong, then the electric force can shift the equilibrium position of the system quite a bit. So, in the new equilibrium position, the amount of stretch, x, of the spring from its natural length, L, can be large. It seems to me that you should expand about the new equilibrium position rather than about L. This will modify your answer for the effective spring constant. You can then see if it affects your final conclusion about the period.
 
  • Like
Likes 1 person
  • #23
ehild
Homework Helper
15,523
1,901
What is the correct way of converting this two body problem in one body problem ?
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.

There are two point masses, m1, m2. Their coordinates are x1 and x2. m1 exerts force f12 on m2 and m2 exerts force f21 on m1. Assume the forces act along the line connecting the masses.

f12=-f21=F

[itex]m_1\ddot x_1=-F[/itex].....................(1)
[itex]m_2\ddot x_2=F[/itex].........................(2)

Add the equations:
[itex]m_1\ddot x_1+m_2\ddot x_2=0[/itex].................(3)

The coordinate of the centre of mass is X
[itex]X=\frac{m_1 x_1+m_2 x_2}{m_1 +m_2}[/itex]
so you can write equation (3) as [itex](m_1+m_2)\ddot X=0[/itex]

The centre of mass moves with constant velocity. You can choose the frame of reference fixed to the CM: Then X=0,

[itex]m_1x_1+m_2x_2=0[/itex].

Divide equation (1) with m1, equation(2) with m2 and subtract them.

[itex]\ddot x_1=-\frac{F}{m_1}[/itex]
[itex]\ddot x_2=\frac{F}{m_2}[/itex]
[itex]\ddot x_2-\ddot x_1=\frac{F}{m_2}+\frac{F}{m_1}[/itex]

Denote x2 -x1 = x
and
[itex]\frac{1}{μ}=\frac{1}{m_1}+\frac{1}{m_2}[/itex]

You get the equation for the relative coordinate x=x2-x1 in terms of the reduced mass μ and the force of interaction F

[tex]μ \ddot x= F[/tex], the same, as in case of a single body of mass μ.

ehild
 
  • Like
Likes 1 person
  • #24
1,540
134
Maybe I'm just being a nuisance
I beg to differ :smile:

but in your equation $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{(x+L)^2}$$ it is not necessarily true that x<< L for small oscillations.
Do you mean x<< x_e for small oscillations ?

If the spring constant is weak while the charge q is strong, then the electric force can shift the equilibrium position of the system quite a bit. So, in the new equilibrium position, the amount of stretch, x, of the spring from its natural length, L, can be large. It seems to me that you should expand about the new equilibrium position rather than about L. This will modify your answer for the effective spring constant. You can then see if it affects your final conclusion about the period.
Please have a look at post#8 .Should I continue with that ?
 
Last edited:
  • #25
ehild
Homework Helper
15,523
1,901
When x<<L , $$\mu \frac{d^2 x}{dt^2} = -kx + k_e\frac{q^2}{L^2}(1-\frac{2x}{L})$$

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

Here effective spring constant ##K = k+\frac{2k_eq^2}{L^3}## and time period is ## T = 2\pi \sqrt{\frac{\mu}{K}}##

Is this correct ?
You should expand around the equilibrium.

ehild
 
Last edited:

Related Threads on Two masses and spring oscillation

  • Last Post
Replies
4
Views
31K
Replies
4
Views
3K
Replies
7
Views
5K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
8K
Replies
12
Views
2K
Replies
5
Views
959
Replies
0
Views
3K
Replies
1
Views
2K
Top