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This has puzzled me quite a bit . To illustrate my doubt we can take a situation where I take a block and an ideal spring and hang them in a vertical line the spring attached to the ceiling .(the spring is in its natural state initially) Now if I were asked to find the maximum elongation of this spring.

My initial approach I equate the forces,so,

$Mg=kx$

Implying , $x=Mg/k$

But using Work energy theorem , $Mgx-(1/2)kx^2=(1/2) m(0)^2-(1/2) m(0)^2$

But this gives $x=2Mg/k$

How come this disparity?

II) Will the conditions of equilibrium points change when the spring is gradually released in one case and suddenly released in other case? Will the elongation change in the above cases?

Nehaa
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Deshna Thunga
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3 Answers3

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Consider the following two cases:

  1. Case 1: The mass is attached to the spring, and you gently extend it to its new equilibrium position. You make sure that it does not oscillate. The mass stays still. Let's call this equilibrium position $x_\text{eq}$.

  2. Case 2: The mass is attached to the spring and released instantaneously. In this case, it extends well past its equilibrium position until it reaches all the way down to some $x_\text{amp}$ where it slows down and gets pulled back, beginning to oscillate (as spring-mass systems do).

The question as it stands is not clear. Which of these two cases are you dealing with? Once you know that, you will know which of the two represents the answer you are interested in. Clearly, the maximum displacement in both cases is not the same.

In the first, the spring is extended until the forces are balanced. This gives you the result $$x_\text{eqm} = \frac{Mg}{k}.$$

In the second, the maximum displacement will be when the mass comes to rest at the extremum of an oscillation. This will be the distance to the equilibrium position plus the amplitude of the oscillation. This is the "furthest" such a spring could extend, and as you've shown, it is given by $$x_\text{amp} = \frac{2 Mg}{k}.$$

Philip
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For an ideal spring:

$$ f(x) = -kx \ \ \forall x$$

so there is no maximum nor minimum extent. Hence: "ideal"

Regarding the "work energy theorem" application: why is the final velocity 0?

You have computed the maximum displacement, which is exactly twice as far as the equilibrium position....because if you drop the mass from $x=0$, it will oscillate around the equilibrium position, $x_0$, with amplitude $\pm x_0$.

JEB
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$Mg=kx \rightarrow $$x=Mg/k$

This is the position where net force is zero and not the velocity.

The mass at that point still has some velocity due to which spring elongates further.

Your second approach is correct.

Bhavay
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