Condition of simple harmonic motion

[kelvin macks]

Homework Statement

I was told that for an object to execute SHM, the x (distance displaced from the spring) can't be greater than e . Why is this so? i can't understand. can someone explain please?

Homework Equations

The Attempt at a Solution

 

[haruspex]

I was told that for an object to execute SHM, the x (distance displaced from the spring) can't be greater than e .

First, you understand that 'e' here does not mean 2.7..., right? It's just the label being used for a variable.
In the text you posted, it stands for the extension of the spring when the mass is hanging at equilibrium. When oscillating, the displacement x is relative to that equilibrium position, so the total extension of the spring is e+x.
I do not see anywhere in the text that says x cannot exceed e, so I assume someone else said this. Suppose it does. In SHM, if the displacement is at times x then at other times it is -x. So if x can exceed e it will also happen that the total extension can go negative. Whether that's a problem depends on the type of spring.
Many springs have an 'open' relaxed state. That is, they are capable of being compressed by some amount c before becoming closed up. For other springs, c = 0.
SHM should occur provided x does not exceed e+c.

 

[kelvin macks]

First, you understand that 'e' here does not mean 2.7..., right? It's just the label being used for a variable.
In the text you posted, it stands for the extension of the spring when the mass is hanging at equilibrium. When oscillating, the displacement x is relative to that equilibrium position, so the total extension of the spring is e+x.
I do not see anywhere in the text that says x cannot exceed e, so I assume someone else said this. Suppose it does. In SHM, if the displacement is at times x then at other times it is -x. So if x can exceed e it will also happen that the total extension can go negative. Whether that's a problem depends on the type of spring.
Many springs have an 'open' relaxed state. That is, they are capable of being compressed by some amount c before becoming closed up. For other springs, c = 0.
SHM should occur provided x does not exceed e+c.

well what happen if x exceed e ?

 

[CWatters]

Haruspex explained that. The spring may stop working as a spring if |x| > |e|

In your diagram..

L₀ = unstretched length of a tension spring.
L₀+e = the length with the mass at rest.
x = displacement from L₀+e

If the vertical displacement x is greater than e the mass will rise to a position above L₀ and the spring may stop working as a spring. Many tension springs don't work in compression.

For example this type doesn't work in compression only tension..

This type will handle compression but only up to a limit..

 

[haruspex]

For example this type doesn't work in compression only tension..

Thanks for posting the two images. They explain it much better than I could in words.
For the first image, it may even be true that it only works as a spring under sufficient tension. It may be effectively under tension before it is expanded at all.