Time Period of SHM: Opposite Charges in Springs

In summary, the two charges will attract and move back and forth until they collide. At that point, the force of the electrostatic attraction will pull them back together.
  • #1
Asphyx820
10
0
1. Homework Statement

Two Springs are present (one just infront of the other). The Spring towards the left has +Q charge and towards the right -Q charge (at their ends).The distance between the two charges is d. The Springs are of length l. Find the Time Period of the Simple Harmonic Motion if the charges are of same mass. ( l > d )

Diagram
(Wall)-->(Spring)-->+Q -Q<--(Spring)<--(Wall)

2. Homework Equations

F(elec)=(k Q^2) / (d^2) where k=(1/4)∏ε
F(Spring)=( Kl )

3. The Attempt at a Solution

I know the above two equations, but can't proceed. Is there any other force too? I can't figure out why will the charges move back again? I'm having two confusions

1) The charges are opposite so they will attract each other. When they reach a certain point they will collide (as l > d ) and move back. Is this the reason why they move back? What other equation do i have to use?

2) Is it the Spring force will pulls the charges back before they collide. But it shouldn't be true as ( l > d ) and electrostatic forces are very strong and spring force cannot overcome it. Am i right? so how should i proceed

Pls help me...
 
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  • #2
You can assume that the system is in equilibrium initially, and then you give a little push to the masses. For a simple harmonic motion, the displacement of the charged masses from their equilibrium positions must be small with respect to the distance between them. Find the time period of small oscillations with this assumption. Do not forget that the springs are connected to two opposite walls, so the sum of the spring lengths and the distance between the masses is constant.

ehild
 
  • #3
But what about the electrstatic force. It constantly changes as the distance between the charges change. how to incorporate that. I know by intergrating the force, but then what to do?
 
  • #4
Do Taylor expansion of the Coulomb force around the equilibrium position and keep the constant and first-order terms.

ehild
 
  • #5
I haven't yet learned Taylor expansion. So is there any other method ?
I actually found this question in a magazine. The answer page was torn. So i don't know the answer too.
I have tried solving it 8-10 times but with no success ! (cant reach the final expansion)
I found this sum interesting so I picked it up
I would be helpful to me if you can solve the sum or give me the equations to be solved
I would be learning things both ways.
 
  • #6
Sorry, I am not allowed to solve problems. I can only help.

It is useful to learn how to calculate with small quantities.
Suppose you have to calculate (1+a)2, where a<<1. Decomposing the square, (1+a)2=1+2a+a2. If a<<1 you can ignore the square of it at approximate (1+a)2≈1+2a. Calculate 1.0012and compare it with 1+2*0.001.

Suppose you have a fraction, 1/(1+q) and q << 1.

1/(1+q) is equal to the sum of the geometric series 1-q+q2-q3+... =1/(1+q)
If q<<1 you can ignore the terms with second or higher power, and use the approximation 1/(1+q)=1-q.
Try to calculate 1/1.001 and compare it with 1-0.001.

Here you have the Coulomb force of form A/(R+Δr)2.
Factor out R: You get ( A/(R2) (1/(1+Δr/R)2.
Assume that Δr/R<<1. Try to apply the approximations above.

ehild
 

1. What is the time period of SHM for opposite charges in springs?

The time period of SHM, or simple harmonic motion, for opposite charges in springs is the amount of time it takes for the system to complete one full cycle of motion. It is determined by the mass of the charges, the spring constant, and the distance between the charges.

2. How is the time period of SHM affected by changes in mass or spring constant?

If the mass of the charges increases, the time period of SHM will also increase. This is because a larger mass requires more force to accelerate, resulting in a slower motion. Similarly, if the spring constant increases, meaning the spring becomes stiffer, the time period of SHM will decrease as the system will oscillate at a faster rate.

3. Can the distance between the charges affect the time period of SHM?

Yes, the distance between the charges can also affect the time period of SHM. As the distance increases, the force between the charges decreases, causing the system to oscillate at a slower rate. Conversely, a decrease in distance will result in a shorter time period.

4. How does the time period of SHM relate to the frequency of oscillation?

The time period of SHM is inversely proportional to the frequency of oscillation. This means that as the time period increases, the frequency decreases, and vice versa. The frequency is the number of cycles per second, while the time period is the duration of one cycle.

5. Is the time period of SHM affected by external factors?

The time period of SHM for opposite charges in springs is not affected by external factors such as air resistance or friction. This is because these forces do not alter the properties of the system, such as mass, spring constant, or distance between the charges, which determine the time period.

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