What is the equation of wave velocity for an extended spring?

Click For Summary
SUMMARY

The wave velocity for a spring is defined by the equation v = √(k / m) x L, where k is the spring constant, m is the mass of the spring, and L is the length of the spring. In scenarios where both ends of the spring are extended and held by an oscillator, the wave velocity can still be derived from this equation, but it is crucial to differentiate between the velocity of the end of the spring and the actual wave velocity. For longitudinal waves, the tension in the spring must be considered, and the formula v = √(T / μ) can be used, where T is the tension and μ is the mass per unit length. Understanding the distinction between displacement and length is essential for accurate calculations.

PREREQUISITES
  • Understanding of spring constants (k) and mass (m)
  • Knowledge of wave mechanics and oscillations
  • Familiarity with tension in elastic materials
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the relationship between tension (T) and wave velocity in springs using the formula v = √(T / μ)
  • Explore the differences between longitudinal and transverse wave propagation in elastic materials
  • Study the effects of mass distribution on wave velocity in springs
  • Learn about potential energy in springs and its relation to oscillation frequency
USEFUL FOR

Physicists, mechanical engineers, and students studying wave mechanics or oscillatory systems will benefit from this discussion, particularly those interested in the dynamics of springs and wave propagation.

null void
Messages
102
Reaction score
1
For a free spring, the wave velocity is v = √(k / m) x L, where L is the displacement of one end right?

And what if I have a spring which both of its end is extended and held by "oscillator" to produce standing wave. In this case how would I find the wave velocity? I see some sources state that the equation i mentioned above is applicable. But wasn't L is the displacement? Or I did get the wrong idea about L ?
 
Physics news on Phys.org
null void said:
For a free spring, the wave velocity is v = √(k / m) x L, where L is the displacement of one end right?

And what if I have a spring which both of its end is extended and held by "oscillator" to produce standing wave. In this case how would I find the wave velocity? I see some sources state that the equation i mentioned above is applicable. But wasn't L is the displacement? Or I did get the wrong idea about L ?

I don't know if this i correct way to approach your problem, but we can calculate the potential energy of the spring from the spring constant and the displacement (L) from PE=\frac {kL^{2}}{2}

We then can then solve the equation PE=mv^2 for the velocity. I think the math is correct but I'm not sure it answers your question.

EDIT: I'm thinking of displacement as the difference in the length of the relaxed spring and the length of the extended or compressed spring.
 
Last edited:
L is probably the length of the spring.

You also need to check exactly what "m" is here. Is it the mass per unit length, or the mass of the whole spring?

If you are talking about axial waves in the spring, the wave velocity is independent of the tension in the spring, unless you are considering very large axial strains (i.e. stretching a floppy spring by a large amount). But to model that effect, you would need to consider the details of how the spring was made, e.g. a coil spring would be different from a flexible elastic band.

If you are talking about transverse waves, the tension in the spring does change the wave velocity - but the formula in your OP doesn't make much sense for transverse waves.
 
AlephZero said:
L is probably the length of the spring.

I defined L as the difference between the relaxed spring and the extended or compressed spring for calculating potential energy (following the OP's example). For a relaxed spring, PE=0. However, the convention may be to use "x" for displacement and L for the length of the relaxed spring. You can extract "velocity" from k, x and m, but I'm not sure how velocity, so calculated, applies in this case.
 
Last edited:
SW VandeCarr said:
I defined L as the difference between the relaxed spring and the extended or compressed spring for calculating potential energy (following the OP's example). For a relaxed spring, PE=0. However, the convention may be to use "x" for displacement and L for the length of the relaxed spring. You can extract "velocity" from k, x and m, but I'm not sure how velocity, so calculated, applies in this case.

OK, but you calculated the potential energy of the spring when it is in equilibrium (and stretched). For the vibration frequency you want the change PE for a small displacement from the equilibrium position.

Reading your post again, I think you are assuming the OP means a light spring with a mass on the end. I was assuming a heavy spring with no additional mass.

But either way, it's not very clear to me what the OP's formula is meant to be. The frequency is √(k/m) rad/sec, with the appropriate definitions of k and m for the physical situation. But to turn that into a wave velocity for a spring of length L, shouldn't there be a 2π somewhere in the formula? And if we assume L is something to do with the amplitude of the wave, it shouldn't be in the formula at all.

Unless "wave velocity" means the velocity of one point on the spring, and not the velocity of the traveling waves in it ... :confused::confused::confused:
 
  • Like
Likes   Reactions: 1 person
v = √(k / m) x L, k is the spring constant , m is the mass of the spring and L is the length of the spring. This formula looks like this one:

v = (+/-)√(k / m) √(A2-x2), where A is the amplitude and x is the displacement from equilibrium point of an oscillating object at the end of a spring.

How to relate this to the wave velocity of an stretched + vibrating spring? And yeah this is not a heavy spring
 
null void said:
v = √(k / m) x L, k is the spring constant , m is the mass of the spring and L is the length of the spring. This formula looks like this one:

v = (+/-)√(k / m) √(A2-x2), where A is the amplitude and x is the displacement from equilibrium point of an oscillating object at the end of a spring.

How to relate this to the wave velocity of an stretched + vibrating spring? And yeah this is not a heavy spring

The second formula contains the variable 'x' (displacement) so it must apply to a stretched (or compressed) spring already if x is non zero. If the formula is correct and you are given k, x, the amplitude and mass, it seems you get a velocity from this formula.

If you measure the tension T (in Newtons) on the spring directly you should be able to use this formula:

v=\sqrt\frac{T}{m}

You are essentially treating the spring like any elastic "string" by knowing the tension. An oscillating mass at the end of the string will produce longitudinal waves and you would use the mean tension since it varies with the oscillations.
 
Last edited:
  • Like
Likes   Reactions: 1 person
so do u know what is the working principle behind this formula ? v = √(k / m) x L, i just can't link how the √(A2-x2) can become the stretched length of spring
 
null void said:
For a free spring, the wave velocity is v = √(k / m) x L, where L is the displacement of one end right?
I am not sure if anyone has pointed this out:

This is NOT the wave velocity. What you wrote is the velocity of the end of the spring.
For a longitudinal wave in a spring the points of the spring may oscillate with velocities given by similar formulas, where L is replaced by displacement from equilibrium position. But this will not be the wave's velocity. They are quite different things.
 
  • Like
Likes   Reactions: 1 person
  • #10
I would like an answer to this question myself. Consider the OP's example of an oscillating suspended weight at the end of a spring. What is the velocity of the waves produced in the spring?

First, are "waves" produced in the spring from the oscillations? It seems the spring would just stretch and recoil by a fixed amount over the whole length of the spring. However, if the spring were compressed over a length 'x' < L, and this compressed segment were released, I could imagine a longitudinal wave propagating along a sufficiently long spring. It seems the easiest way to calculate the velocity of this wave is to use the tension ("T") formula:

v=\sqrt {\frac{T}{\mu}}

Tension has the dimension of force and \mu is the mass per unit length of the spring. Is this correct or is it at least correct regardless of how such a wave is generated?

EDiT: The above formula can also be expressed in terms of the spring constant as T=kx.
 
Last edited:
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 27 ·
Replies
27
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K