Hooke's law for vibrating massive spring

In summary, the tension of a vibrating massive spring depends on both the elongation and the acceleration. However, the acceleration is multiplied by the mass of the spring, so if the spring is massless, the tension is reduced to T = kx.
  • #1
bgq
162
0
Hello,

Just for curiosity...
Is Hooke's law valid for a vibrating massive spring ?
I have done some calculations using both Newton's 2nd Law and the conservation of energy to a horizontal swinging spring connected to a small block in the absence of any friction. I have found that the tension of the spring depends on both the elongation and the acceleration.
However, the acceleration is multiplied by the mass of the spring, so if the spring is massless, the tension is reduced to T = kx.

Here is the outline of my work:

>> I have written the expression of the mechanical energy of the system (block-spring), and then set the derivative to zero. I concluded at the end that I can neglect the mass of the spring if I assume the mass of the block is M + m/3 where m is the mass of the spring and M is the mass of the block. I checked this out on the internet, and I found this conclusion true.

>> Next I have applied Newton's 2nd Law:
For a massless spring: T = Ma
For a massive spring, the tension is T':
T' = (M + m/3)a (Neglect the mass of the spring and add its third to the block)
T' = Ma + ma/3
T' = T + ma/3
T' = -kx + ma/3 (Note that the tension depends on the acceleration).

Is my work correct?

Thank you.
 
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  • #2
Hooks Law is an idealization - it is a useful approximation for many real springs and spring-like situations. So the short answer is "no" - it is not strictly true for any real-World situation.

I can neglect the mass of the spring if I assume the mass of the block is M + m/3 where m is the mass of the spring and M is the mass of the block.
You are saying that you can neglect the mass of the spring if M = M + m/3 ??
Do you mean that if you have a spring mass m and block mass M system, you can model it by an ideal massless spring and a block of mass M+m/3 and get physically the same result for something? (What is it that needs to be the same? Acceleration? The relationship between the restoring force and the extension?)

I checked this out on the internet, and I found this conclusion true.
If it is on the internet then it must be true!
(But JIC: please provide the URL.)

Is my work correct?
Can't tell. You need to be more careful.
 
  • #3
I don't remember the URL, but here is my work.

Let M be the mass of the spring, L is its length at any time t, and V is the speed of its free end at the time t. Let x be the distance from the fixed end of the spring to any infinitesimal part of the spring, and v is the speed of this part. Let λ be the linear mass density of the spring.

Assume that the speed of any infinitesimal part of the spring is proportion to x, then v = xV/L.

The kinetic energy of an infinitesimal part is: dK = 1/2 dm v2 = 1/2 λdx (xV/L)2 = 1/2 M/L (V/L)2 x2dx
The kinetic energy of the spring is:
K = ∫dK (from 0 to L) = (MV2)/(2L3) ∫x2dx (from 0 to L) = (MV2)/(2L3) [x3/3] (0 to L) = 1/6 M V2

Now the total kinetic energy of the system spring-block: K = 1/6 M V2 + 1/2 m V2 = 1/2(m+M/3)V2 which is totally equivalent to consider the spring massless but adding the third of its mass of the block.

In what follows x is the elongation (or compression) of the spring.

The total mechanical energy of the system Spring-Block is: E = K + U = 1/2(m+M/3)V2 + 1/2kx2

Since the mechanical energy is constant then its derivative is zero, so

(m + M/3) x'' + kx = 0 then mx'' + Mx''/3 + kx = 0
-Mx''/3 - kx = mx''
-Ma/3 - kx = ma
The right side is the mass multiplied by the acceleration of the block, so the left side should equal to the net external force acting on the block which is the tension of the spring.

Therefore the magnitude of the tension of the spring is T = - kx - Ma/3.
 
Last edited:
  • #4
The relation for U is only if the spring obeys Hooks Law.
 
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Likes bgq
  • #5
Simon Bridge said:
The relation for U is only if the spring obeys Hooks Law.

Oh, I missed this point. Thanks a lot. I will try to do it again.
 
  • #6
Good luck.
 

Related to Hooke's law for vibrating massive spring

What is Hooke's Law for vibrating massive spring?

Hooke's Law for vibrating massive spring is a physical law that describes the relationship between the force applied to a spring and the resulting displacement of the spring from its equilibrium position. It states that the force required to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

How does the spring constant affect the behavior of the spring?

The spring constant, k, determines how stiff or soft the spring is. A higher spring constant means that the spring is stiffer and requires more force to stretch or compress, while a lower spring constant means the spring is softer and requires less force to stretch or compress.

What is the significance of the negative sign in Hooke's Law equation?

The negative sign in the Hooke's Law equation signifies that the force and the displacement are in opposite directions, meaning that the force applied to the spring is always directed towards the equilibrium position.

Can Hooke's Law be applied to all types of springs?

Hooke's Law can be applied to most springs as long as they are not stretched or compressed beyond their elastic limit. This means that the spring is not permanently deformed and can return to its original shape once the force is removed.

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