I have been reading Free Vibration of Multi Degree of Freedom Systems and have got the following clarifications in regard to it:(adsbygoogle = window.adsbygoogle || []).push({});

1) When a free vibration is initiated with an initial displacement, the multi degree of freedom system may not undergo simple harmonic motion, right?

It will undergo simple harmonic motion only if the free vibration is initiated with a proper distribution of displacements in various degrees of freedom, right? These characteristic deflected shapes wherein the system vibrates in simple harmonic motion is are called as “Natural Modes of Vibration”, right?

Thus, in practical problems the system may not vibrate in its natural modes of vibration at all, right?

Then why we have such detailed studies in various texts for the systems vibrating in their natural modes of vibration?

2) The free vibration of an undamped system is for a multi degree of freedom system in one of its natural vibrating modes is described as:

u (t) = qn(t)Øn

Where, the deflected shape Øn does not vary with time .The time variation of the displacements is described by the simple harmonic function

qn(t)= An cos ωnt + Bn sin ωnt

What is the justification of the relation

u (t) = qn(t)Øn

3) Substituting the relation:

U (t) = Øn (An cos ωnt + Bn sin ωnt)

In the differential equation

Inertial force (mass x acceleration)+ Elastic force(k x u)=0

(Considering undamped system and free vibration)

and further simplifying gives the frequency equation

Determinant [ k - ωn2m] Øn = 0

Solution of the above equation gives “N” values of frequency ωn , each value of ωn corresponds to a particular natural mode , right?

However, the main equation u (t) = qn(t)Øn was written for the nth mode of vibration only , however simplifying the same we get frequencies of each mode. Right?

How was that possible.How are we justified in saying that each value of ωn corresponds to a particular natural mode

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# Homework Help: Multi Degree of Freedom System

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