Two masses and spring oscillation

In summary, the question asks whether charging the masses with electric charges affects the oscillation period of an insulated spring and mass system. The correct approach is to use the concept of reduced mass and convert the two-body problem into a one-body problem, with the relative separation 'x' as the x-coordinate of the second mass. The differential equation for this problem is given by $\mu \frac{d^2 x}{dt^2} = -kx$, and for the charged masses, the electric force can be added and expanded to first order to find the effective spring constant. The correct approach is to use the concept of reduced mass and convert the two-body problem into a one-body problem, with the relative separation 'x' as the
  • #1
Tanya Sharma
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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 .
 

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  • #2
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:
 
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  • #3
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
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
NascentOxygen said:
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
Tanya Sharma said:
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
dauto said:
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
ehild said:
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 ?
 
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  • #9
Tanya Sharma said:
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
Do you know about reduced mass and converting a two-body problem into a one-body problem?
 
  • #11
Hello Vela

vela said:
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
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
vela said:
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
Looks good.
 
  • #15
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
 
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  • #16
Right idea, but you made some algebra mistakes.
 
  • #17
OK..I have fixed the errors .Does post#15 make sense ?
 
  • #18
Yup.
 
  • #19
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 ?
 
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  • #20
Yes.

ehild
 
  • #21
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
Tanya Sharma said:
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.
 
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  • #23
Tanya Sharma said:
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
 
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  • #24
TSny said:
Maybe I'm just being a nuisance

I beg to differ :smile:

TSny said:
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 ?

TSny said:
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 ?
 
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  • #25
Tanya Sharma said:
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
 
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  • #26
ehild said:
Yes, and write the differential equation for x-xe.

ehild

ehild said:
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
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
ehild said:
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
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
ehild said:
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
Expand f(x)=keq2/(x+L)2 around xe.

ehild
 
  • #32
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
Tanya Sharma said:
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
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
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 .

ehild said:
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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<h2>1. What is a mass-spring system?</h2><p>A mass-spring system is a physical system composed of two masses connected by a spring. The mass-spring system exhibits oscillatory motion when the spring is compressed or stretched, causing the masses to move back and forth around their equilibrium position.</p><h2>2. What is the equation for the motion of a mass-spring system?</h2><p>The equation for the motion of a mass-spring system is given by <strong>F = -kx</strong>, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.</p><h2>3. How does the mass affect the oscillation of a mass-spring system?</h2><p>The mass of the system affects the period and frequency of the oscillation. A larger mass will result in a longer period and lower frequency, while a smaller mass will result in a shorter period and higher frequency.</p><h2>4. What factors affect the amplitude of a mass-spring system?</h2><p>The amplitude of a mass-spring system is affected by the initial displacement, the mass, and the spring constant. A larger initial displacement, mass, or spring constant will result in a larger amplitude of oscillation.</p><h2>5. How does the spring constant affect the oscillation of a mass-spring system?</h2><p>The spring constant determines the stiffness of the spring and affects the period and frequency of the oscillation. A higher spring constant will result in a shorter period and higher frequency, while a lower spring constant will result in a longer period and lower frequency.</p>

1. What is a mass-spring system?

A mass-spring system is a physical system composed of two masses connected by a spring. The mass-spring system exhibits oscillatory motion when the spring is compressed or stretched, causing the masses to move back and forth around their equilibrium position.

2. What is the equation for the motion of a mass-spring system?

The equation for the motion of a mass-spring system is given by F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

3. How does the mass affect the oscillation of a mass-spring system?

The mass of the system affects the period and frequency of the oscillation. A larger mass will result in a longer period and lower frequency, while a smaller mass will result in a shorter period and higher frequency.

4. What factors affect the amplitude of a mass-spring system?

The amplitude of a mass-spring system is affected by the initial displacement, the mass, and the spring constant. A larger initial displacement, mass, or spring constant will result in a larger amplitude of oscillation.

5. How does the spring constant affect the oscillation of a mass-spring system?

The spring constant determines the stiffness of the spring and affects the period and frequency of the oscillation. A higher spring constant will result in a shorter period and higher frequency, while a lower spring constant will result in a longer period and lower frequency.

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