A rod of length L and mass m is clamped at one end and attached axially to a spring of stiffness k at the other end. Determine the natural frequencies for the longitudinal vibration of the rod?

Question

experimentx · Answer

The  time period of the harmonic spring oscillator is given by

$$ f = 2\pi \sqrt{k/m} $$

No use the relation
$$\omega = \frac{1}{T}$$ to get the frequecncy

anonymous · Answer

hey @experimentX : dont u think that instead of m something else would be there for a rod?...after all it is not a point mass..

experimentx · Answer

hahahah ... i was going to type "usually we take center of mass for it. However there will always be an error."

Usually i never read questions, I thought it would be a sphere. Let's see what we can do.

anonymous · Answer

As i see it...when the spring will push or pull the rod it will move as well as get compressed or elongated..so some of the energy that were in vibration will get stored inside the rod as potential (strain) energy..so it may affect the natural frequency..

experimentx · Answer

Well if the rod is not rigid then ...

experimentx · Answer

still there will be an error because the condition you say applied to spring ... and spring is also the part of the system.

anonymous · Answer

mass m is clamped at 'one end' and attached axially to a spring of stiffness k at the 'other end'...oh drat..i thought that the spring was clamped..

anonymous · Answer

now how is it going to vibrate now with rod clamped?

experimentx · Answer

??? I am lost!! can you draw the picture please??

anonymous · Answer

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