Example of oscillations
Example: A thin rod of length L and area of cross-section S is pivoted at its lowest point P inside a stationary, homogeneous and non viscous liquid (fig.1). The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density d, of the material of the rod is smaller than the density d2 of the liquid. The rod is displaced by a small angle θ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.
Solution:
When the rod is displaced slightly through an angle θ with the vertical Fig.2, the effective upward force acting at the middle point G of the rod is:
F= upward thrust – weight of rod
= weight of liquid displaced – weight of rod
=LSd2g – LSd1g =LSg(d2-d1)
Moment of the couple (i.e. torque) tending to bring the rod to its original position is:
Torque = F × KG
= LSg(d2-d1) ×L sinθ/2
= [Sg(d2-d1)L2/2] θ …………………………(i)
Θ is small
So torque is proportional to θ. As this torque is restoring one, hence if the rod is left free it will execute SHM. Compare (I)
Torque = k θ, we have
Spring factor = Sg(d2-d1)L2/2
Here, inertia factor = moment of inertia of the rod about axis at P
= ML2/2+M(L/2)2 = ML2/3 =L3Sd1/3
As angular frequency in S.H.M. is
ω= √(spring factor/inertia factor)=√[3g(d2-d1)/2Ld1]