# Mass and vibration frequency

Deepesh
TL;DR Summary
This is regarding Stiffness and mass included in Vibrations subject and natural frequency of an object
My query here is,
Suppose there is a 2 kg mass
To oscillate it/vibrate it, it will take some force and it will have some natural frequency
Now I increase the mass to 5 kg
so to vibrate it, won't it take more force and so at the end, won't the natural frequency of the object increase? as its more heavy and requires more force to vibrate it?

Homework Helper
Hello Deepesh, !

Summary:: This is regarding Stiffness and mass included in Vibrations subject and natural frequency of an object

My query here is,
Suppose there is a 2 kg mass
To oscillate it/vibrate it, it will take some force and it will have some natural frequency
Now I increase the mass to 5 kg
so to vibrate it, won't it take more force and so at the end, won't the natural frequency of the object increase? as its more heavy and requires more force to vibrate it?

Vibrations occur when there is an equilibrium situation and some restoring force that works towards this equilibrium. The simplest way to describe this in physics (mathematics) is $$m\ddot x + kx = 0$$with as solutions ##x = A\sin(\omega t + \phi)##. Substitution shows ##\omega^2 = \displaystyle {k\over m}##. In other words: frequency decreases when only ##m## increases.

And it increases when only ##k## (related to your stiffness) increases.

Lnewqban
Deepesh
If we can keep aside formulae for a second?
If you can explain it to me like what happens there practically?
If mass is increasing, isn't making difficult for the force to move/vibrate that body and thus shifting the natural frequency up?

Homework Helper
Correct. Inertia wins.

Deepesh
Now if we come back to simple formula for natural frequency i.e omega= underoot (k/m)

If we increase mass here, the N.F would increase.
Inverse proportion

How does this relate to practically that we just discussed?

Thanks

Homework Helper
I don't understand the question

Homework Helper
Gold Member
Now if we come back to simple formula for natural frequency i.e omega= underoot (k/m)

If we increase mass here, the N.F would increase.
Inverse proportion

How does this relate to practically that we just discussed?

Thanks
If you replace mass2 with mass2 in that equation, being mass2 > mass1, the value of the natural frequency (1/seconds) decreases and the value of the period increases (if keeping amplitude and phase the same).
Period = 2 pi / natural frequency

It is only the Second law of Newton.
Taking a spring-mass harmonic vibration for example:
Let's stop when the both masses (small and big) are located at maximum displacement from the equilibrium position.
Both are being pulled towards that equilibrium position by forces of same magnitude, as both masses are attached to similar springs (k) suffering same deformation (x).
Basically, each mass is "loaded" at maximum acceleration and minimum speed.

Now, what happens when we release both masses at the same time?

Spring1 force = spring 2 force = mass1 * acceleration 1 = mass2 * acceleration 2

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