Free Vibration of Spring system with two DOF

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For a two-degree-of-freedom spring system undergoing free vibration, the determinant of the coefficients in the simultaneous equations must be zero to find non-trivial solutions for the displacements X1 and X2. This condition arises because, without external forces, the equations equate to zero, leading to a trivial solution if the determinant is non-zero. The special values of ω that make the determinant zero correspond to the natural frequencies of the system, which are critical for understanding its vibrational behavior. The discussion highlights the importance of these frequencies in analyzing the system's dynamics. Understanding why the determinant must be zero is essential for grasping the underlying mechanics of free vibration in multi-degree systems.
nerak99
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In some questions I am doing, you set of a pair of simultaneous equations and in the notes we have that . For a non trivial solution of X1 and X2,the determinant of
coefficients of X1 and X2 must be zero.

An equation might typically look like this (m \omega^2 +k_1)x_1 +k_2 x_2=0

Why must the determinant be zero? When solving SEs in general using matrices, the determinant must be non-zero?
 
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For free vibtration, there are no external forces appplied to the system, so the right hand sides of the equations are 0.

If the determinant is non-zero, the only solution is ##x_1 = x_2 = 0## which is not very interesting!

The determinant is only zero for "special" values of ##\omega##, and these are the frequencies at which the system can vibrate.
 
Thanks for that. I still don't understand "why" but that is v helpfule.
 
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