How Does Doubling the Mass Affect the Frequency of a Spring System?

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Doubling the mass in a spring system affects the frequency of oscillation, as described by the equation ω = √(k/m). The correct interpretation of the system involves understanding that when two identical masses are used, the effective mass becomes 2m, leading to a frequency of √(k/2m). However, the spring's behavior during oscillation requires considering the relative movements of both masses; if one mass moves x to the right, the other must move x to the left to maintain equilibrium. This results in the spring stretching by a total distance related to the displacements of both masses. Understanding these dynamics is crucial for solving problems related to oscillatory motion in systems with multiple masses.
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Homework Statement


https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf
#16


Homework Equations


ω = √(k/m)


The Attempt at a Solution


So in this problem, I thought that given two identical masses on each side (that exert let's say Force F each), the net tension is not 2F, but F. This the force that is resisted is F = - kx, not twice that. I thought that there was nothing special about k, but apparently I'm wrong. Also, I thought given two objects, the mass is 2m; yielding the following answer:
√(k/2m) - The wrong answer.

The correct one is :
√(2k/m)
 
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Imagine the blocks sitting at rest with the spring unstretched so each block is at its equilibrium position. Think about how you would start the system oscillating so that each mass oscillates about its equilibrium position with the midpoint of the spring always staying at rest.

If at some moment of the oscillation the block on the right is a distance x to the right of its equilibrium position, where is the other block in relation to its equilibrium position? How much is the spring stretched in terms of x? How would you write the force that the spring exerts on the right-hand block in terms of x?
 
Well if A is the amplitude, I would think that as one block is X to the right, the other would be A-X to the left. The spring is stretched by A-X + X = A? The force that the spring exerts on the right-hand block I'm not sure how to write.
 
If one block has moved 0.1 m from its equilbrium position, how far has the other block moved from its equilibrium position? How much has the spring stretched?
 
The other block compresses .1m?
 
SignaturePF said:
The other block compresses .1m?

Yes, for the .1 m. Not sure about the interpretation of "compresses". Consider the figure below which shows the equilibrium position of each mass. If at some instant block B happens to be displaced a distance x to the right of its equilibrium position as shown, then where is block A at this instant in relation to its equilibrium position?
 

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To the right of its EP.
 
SignaturePF said:
To the right of its EP.

No. If both blocks move to the right by the same distance, then the whole system simply displaces to the right with no change in the amount of stretch of the spring. That would not correspond to oscillatory motion of the system.
 
Whoops. Where can I study oscillatory motion with two masses - my text is no help.

Must it move x to the left?
 
  • #10
SignaturePF said:
Must it move x to the left?

Yes, it must move to the left.

Imagine starting the system oscillating by simultaneously pulling block A to the left a certain distance and block B to the right the same distance and letting them go at the same instant. Can you visualize what the system will do after that?
 
  • #11
They will oscillate between getting relatively close to relatively far away
 
  • #12
So, if block B has displaced a distance x to the right of its equilibrium position, block A will have displaced a distance x to the left of its equilibrium position. How much has the spring stretched from its natural length? (Express in terms of x.)
 
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