00:01
So given in the question, mass of a uniform rod is 3 kg.
00:08
It is pivoted at the point a.
00:10
At the other end c, there is a circular solid mass, 4 of 4 cages.
00:18
At the point b is hinged by a spring arrangement.
00:23
Point c is also hinged by another spring arrangement.
00:31
It has been asked to find the natural system frequency, and the time taken by the system to complete one oscillation, that is its time period.
00:46
Now the theory, the equivalent spring constant of two springs having coefficient of elasticity or the spring constant, k1 and k2, connected in parallel, to be k equivalent is k1 plus k2.
01:05
Similarly, if the two springs are connected in series, then the spring constant would be 1 by k equivalent, that is 1 by k1 plus 1 by k2.
01:21
If we displace the rod by a small angle theta, it will start executing angular simple harmonic motion defined by the relation theta double down plus omega square theta is equal to 0.
01:38
Where omega is the angular frequency and it is defined as 2 pi by t where t is the time period of oscillation, it is this angular simple harmonic motion, which arises due to the net torque acting on the system, which is defined as i theta double dot, that is equal to draw.
01:56
The torque comes out due to these two spring forces acting about at the point b and at the point c as a result of extension in this spring and a compression in this spring arrangement due to this angular displacement.
02:16
So let us now solve the problem.
02:20
If we replace the spring constants or the spring arrangements hinged at point b and point c with individual springs having equivalent spring constant k and k prime...