A uniform rod of mass `m=1.5 kg` suspended by two identical threads `l=90 cm` in length ( figure ) was turned through a small angle about the vertical axis passing through its middle point `C`. The threads deviated in the process through an angle `alpha=5.0^(@)`. Then the rod was released to start performing small oscillations.
Find`:`
`(a)` the oscillation period,
(b) the rod's oscillation energy.

Explanation

`(a)` Let us locate the sytem when the threads are deviated through an angle `alpha^(') lt alpha`, during the oscillations of the system `(` figure`)`. From the conservation of mechannical energy of the system `: `
`(1)/(2)(mL^(2))/(12)theta^(2)+mgl(1- cos alpha^('))=` constante `(1)`
where `L` is the length of the rod, `theta` is the angular deviation of the rod from its equilibrium position i.e.
`theta=0`
Differentiating Eqn. (1) w.r.t. time
`(1)/(2)(mL^(2))/(12)2 dot(theta)ddot(theta)=mgl sin alpha^(')dot(alpha^('))=0`
So, `(L^(2))/(12) dot(theta)ddot(theta)+glalpha^(')dot(alpha^('))=0 ` ( for small `alpha^('), sini alpha^(')=alpha^(')) ....(2)`
But from the Figure,
`(L)/(2) theta=l alpha^(')` or `a^(')=(L)/(2l) theta`
So,` dot(alpha^('))=(L)/(3l)dot(theta)`
Putting these values of `alpha^(')` and `(dalpha^('))/(dt)` in Eqn. `(2)` we get
`(d^(2)theta)/(dt^(2))=-(3 g)/(l)theta`
Thus the sought time period
`T=(2pi)/(omega_(0))=2pisqrt((l)/(3g))`
`(b)` The sought oscillation energy
`E=U_("extreme")=mgl(1-cos alpha)=mgl2 sin ^(2)((alpha)/(2))`
`=mgl2(alpha^(2))/(4)=(mglalpha^(2))/(2)` ( because for small angle `sin theta ~= theta)`