Simple harmonic motion springs

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The amplitude of oscillation in a spring-mass system is determined solely by the additional displacement (Y) from the equilibrium position, rather than the total displacement (X + Y). This is because simple harmonic motion (SHM) relies on the restoring force being proportional to the displacement from the equilibrium position. When the mass is displaced, the spring's restoring force acts to return it to this equilibrium, which is defined by the position after the initial stretch (X). Even if the spring is partially stretched when unloaded, the restoring force remains consistent, leading to the same oscillation period. Therefore, the amplitude of oscillation is only influenced by the displacement from the new equilibrium position.
Jas
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I have a spring with mass M attached, and leave it at equilibrium. Then I displace it some more by stretching it down a bit more. Displacement due to the mass= X, displacement due to stretching it even more=Y.

Why isn't the amplitude of oscillation= X+Y, but is only actually only Y? This is really confusing me!

Relevant equations: (L is the natural length of the spring, and x is the extension, λ is the modulus of elasticity
Tension=λx/L
Energy stored=λx^2/2L
 
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Jas said:
I have a spring with mass M attached, and leave it at equilibrium. Then I displace it some more by stretching it down a bit more. Displacement due to the mass= X, displacement due to stretching it even more=Y.

Why isn't the amplitude of oscillation= X+Y, but is only actually only Y? This is really confusing me!

Relevant equations: (L is the natural length of the spring, and x is the extension, λ is the modulus of elasticity
Tension=λx/L
Energy stored=λx^2/2L
What if ##Y = 0##?
 
PeroK said:
What if ##Y = 0##?
Then it would just stay in equillibrium
 
Jas said:
Then it would just stay in equillibrium

But you would have it oscillate with amplitude ##X+ Y = X##.
 
Jas said:
Why isn't the amplitude of oscillation= X+Y, but is only actually only Y? This is really confusing me!
The nature of SHM is that the restoring force is proportional to the displacement from the equilibrium (or rest) position. Y will be positive or negative, depending on whether the mass is above or below position X. We, of course, assume that the k of the spring is the same over the whole range of positions of the mass.
Here's another idea. Assuming the spring is partly open went unloaded (easy to arrange by giving it a good stretching!) Then imagine the spring and mass are laid on a frictionless horizontal surface ( dry, clean ice). The rest position of the mass will now be different, with respect to the fixing of the spring but the restoring force per cm of displacement will still be the same. So the period will be exactly the same as when the mass is hanging down. The amplitude will still be the maximum displacement relative to the rest position.
 

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