So, this is a post regarding a general result which is of a lot of help if remembered in Mechanics problems. The derivation is of how to calculative effective mass of block and spring attached if spring is massive.

Let's consider the mass of spring to be \(m\).

So let's begin,

So let us take a small element of mass \(dm\) of the spring, so total energy can be calculated by adding all the length elements' kinetic energy, and requires the following integral:

\(E = \int \dfrac{1}{2}u^{2} dm\)

Since spring is uniform,

\(dm = (\dfrac{dy}{L})m\)

\(E = \int_{0}^{L} \dfrac{1}{2}u^{2}(\dfrac{dy}{L})m\)

The velocity of each mass element of the spring is directly proportional to its length, i.e.

\(u = (\dfrac {vy}{L})\)

\(E = \dfrac{1}{2} \dfrac{m}{L} \int_{0}^{L} (\dfrac{vy}{L})^{2} dy\)

\(E= \dfrac{1}{2} \dfrac{m}{3} v^{2}\)

Comparing to the expected original kinetic energy formula,

\(K.E. = \dfrac{1}{2}mv^{2}\)

we can conclude that effective mass of spring in this case is \(\dfrac{m}{3}\)

The EFFECTIVE MASS of the spring in a spring-mass system when using an ideal spring of uniform linear density is \(\dfrac{1}{3}\) of the mass of the spring.

And we're done.

## Comments

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TopNewesthow we came to know that velocity of each mass element of the spring is directly proportional to its length???? – Nikhil Tanwar · 2 years, 3 months ago

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– Ritvik Choudhary · 2 years, 3 months ago

It's true that this result is based only on the assumption that velocity of each mass element of the spring is directly proportional to its length, but since above is technique used for solving certain problems in classical mechanics where you can generally take the above assumption as valid.Log in to reply

– Nikhil Tanwar · 2 years, 3 months ago

Thanks RitvikLog in to reply

– Ritvik Choudhary · 2 years, 3 months ago

Glad to help :)Log in to reply

This is only True when we assume Liniar relation of velocity of massive spring ! So it is not the genral result ! – Karan Shekhawat · 2 years, 3 months ago

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