# Why are strings and springs often considered massless?

Ok, I know this type of question is already asked, but in every question I have seen, there is no answer to the question that I am asking right now, like people don't particular focus on the question asked (particularly this question) and they go out of context, that is the reason for me to ask this question i.e to get a clear and valid (satisfying) answer.

So, I have often seen, that it is assumed that strings and springs are considered to be massless, but never got the reason, so as to why this assumption is applied.

For the string:

For strings I have heard that, the reason it is considered massless is because,in order to keep the tension at every point on the string the same.

For the spring:

For the spring part I have heard that,it is considered massless because, if we had to take into account the mass of the spring, then additional "wave propagation" effects in the spring would also have to be taken into account, and the analysis would be more complicated. Also another part of assuming that the spring is mass-less is that the spring force at each point on the spring will be the same. (If the wave propagation part is right then kindly elaborate it)

If what I have heard about the springs and strings is true then please let me know, that I am right, if not then please provide the correct knowledge.It would be very helpful if the answer is divided into 2 parts i.e for spring and for string,also if possible, then I would also like to know what would had happened if the masses of the springs and strings have been taken into consideration.

• If you're an undergrad student, it's because if they aren't then it's too much for you to handle. Commented Sep 13, 2020 at 21:18
• Sir can you at least name the phenomenon or maths involved if the masses where considered.If according to you it's so hard the just tell me the names and if you want to you can elaborate also. Commented Sep 13, 2020 at 21:21
• I can't since I never did graduate physics. Another point to consider is that if you treat the spring or string as having mass you are also probably treating the weight as a point. You can help account for mass of the spring or string by adding it in and adjusting the COG of the assembly but this adds little meaningful content to the problem. As soon as you stop treating things as a point (and it seems like you would have to if you want to take into account mass of the string or spring in any meaningful way) then it sounds like it becomes a finite element analysis situtation. Commented Sep 13, 2020 at 21:25
• @DKNguyen None of this is "too hard for undergrad students to handle," but physicists (compared with engineers) are not so interested in real-world situations. The OP might want to look at the vibration of a cantilever beam, as an simple example of a system with a continuous mass distribution (and finite element analysis is not required to solve it). Commented Sep 13, 2020 at 22:37

The classes of problems involving springs are different from the problems involving strings, so you are really asking two questions here. The math involved depends on the problem you are trying to solve. If you want to explore strings with mass, you can search the web for Catenary as one example.

If you have the software to read a Mathematica file you can check out this address to get a Mathematica notebook that solves a vibration problem with a spring with mass:

https://library.wolfram.com/infocenter/MathSource/7773/

Generally ignoring the mass of a spring is a very good approximation if the other masses of a system are substantially greater than the mass of the spring. For instance the masses of the springs supporting a car are much less than the mass of the car, and it would add unnecessary complication to take the mass of the springs into account. Often the same is true for strings or cables.

• IMO there's only one question. The question is not about strings or springs. The question is about the principle of constructing idealized models of physical systems. Commented Sep 14, 2020 at 0:58
• @SolomonSlow I beg to differ since the OP requests the answer be in two parts, one for springs and one for strings. Commented Sep 14, 2020 at 5:55
• @BillWatts thank you sir,it helped Commented Sep 14, 2020 at 8:59

Both reasons you gave are correct.

The concept of a spring is to deform linearly with tension, so it can be more easily modeled as a rod of constant section $$A$$, modulus of elasticity $$E$$, lenght $$L$$, density $$\rho$$ (if it is has mass). I consider here only uniaxial deformation.

If a small element at the position $$x$$ with a length $$\Delta x$$ is selected, the net force acting on it must be the product of its mass and acceleration:

$$F_R - F_L = \rho A\Delta x a$$

The force at left of the element is $$F_L = \sigma(x)A$$
The force at right of the element is $$F_R = \sigma(x+\Delta x)A$$

$$A(\sigma(x+\Delta x) - \sigma(x)) = \rho A\Delta x a$$

Dividing by $$A\Delta x$$ and entering with the definition of acceleration, where $$u$$ is the elastic displacement of the element:

$$\frac{\sigma(x+\Delta x) - \sigma(x)}{\Delta x} = \rho \frac{\partial^2u}{\partial t^2}$$

The left side, when $$\Delta x$$ goes to zero is the derivative:

$$\frac{\partial \sigma}{\partial x} = \rho \frac{\partial^2u}{\partial t^2}$$

Here it can be noted that if the rod is massless, $$\rho = 0$$ and the tension is constant along the rod.

As the rod is linear elastic, tension is proportional to deformation:

$$\sigma = E\epsilon = E\frac{\partial u}{\partial x}$$

Substituting in the previous expression, we got a wave equation:

$$\frac{\partial^2 u}{\partial x^2} = \frac{\rho}{E} \frac{\partial^2u}{\partial t^2}$$

The difference between a string and a spring, if the problem is uniaxial, is that a string can not be compressed, and the deformation is negligible. But the same conclusions is valid for the tensile tension be constant along it, if it is massless.

• Thank you sir,you are amazing Commented Sep 14, 2020 at 10:51
• Sir can you tell me what is d^2(u)/d(x)^2 called.... Commented Sep 14, 2020 at 15:00
• I don't know if it has a standard name. It can be called strain gradient. Commented Sep 14, 2020 at 16:10
• Hmm,also its unit is coming out to be (meter)^-1,so I thought of it as a wave number Commented Sep 14, 2020 at 16:12

when deriving the equations of motion for a dynamical system, ignoring the masses of springs and strings will furnish an accurate and useful approximation of the system for those cases where the excitation frequencies are lower than or equal to the primary resonances of the system. This approximation is far easier to explicitly solve for than the case where those masses need to be included.

In the vast majority of dynamic systems modeling tasks, this approximation yields a satisfactory accounting of the system, and the analysis of the system stops there.

• Honestly speaking,I think you are right about what you are saying sir,but I am not able to get it. Commented Sep 14, 2020 at 9:04