A thin uniform plate shaped as an equlateral triangle with a height `h` performs smalll oscillations about the horizontal axis coinciding with one of its sides. Find the oscillation period and the reduced length of the given pendulum.

Explanation

Consider the moment of inertia of the triangular plate about `AB` .
`I=int int x^(2)dm = int int x^(2)rho d x d y `
`=int _(0)^(h)x^(2)rho d x (h-x)/(h). (2h)/(sqrt(3))=int _(0)^(h)x^(2)(2rho)/(sqrt(3))(h-x)dx`
`=(2 rho )/(sqrt(3))((h^(4))/(3)-(h^(4))/(4))=(rhoh^(4))/(6 sqrt(3))=(mh^(2))/(6)`
On using the area of triangle `Delta ABC=(h^(2))/(sqrt(3))` and `m=rho Delta`.
Thus K.E. `=(1)/(2)(mh^(2))/(6)dot(theta^(2)`
`P.E. =mg((h)/(3))(1-cos theta)=(1)/(2)mgh (theta^(2))/(3)`
Here `theta` is the angle that the instantaneous plane of the plate makes with the equilibrium position which is vertical. `(` The plate rotates as a rigid body `)`
Thus `E=(1)/(2)(mh^(2))/(6)dot(theta^(2))+(1)/(2)(mgh)/(3)theta^(2)`
Hence `omega^(2)=(2g)/(h)=(mgh)/(3)//(mh^(2))/(6)`
So, `T=2pi sqrt((h)/(2g))=pisqrt((2h)/(g))`. and `l_("reduced")=h//2`.