Find the time dependece of the angle of deviation of a mathematical pendulum `80 cm` in length if at the initial moment the pendulum.
(a) was deviated through the angle `3.0` and then set free without push,
`(b)` was in the equlibrium position and its lower end was imparted the horizontal velcoity `0.22m//s,`
(c ) was deviated throught the angle `30^(@)` and its lower end was imparted the velocity `0.22m//s` derected toward the equilibrium position.

Explanation

The natural angular frequency of a mathematical pendulum equals `omega_(0)sqrt(g//l)`
`(a)` We have the solution of S.H.M. equation in angular form `: `
`theta=theta_(m)cos (omega_(0)t+alpha)`
If at the initial moment `i.e.` at `t=0, theta=theta_(m)` than` alpha=0`
Thus the above equation takes the form
`theta=theta_(m)cos omega_(0)t`
`=theta_(m)cossqrt((g)/(l))t=3^(@)cos sqrt((9.8)/(0.8))t`
Thus `theta=3^(@) cos 3.5 t`
`(b)` The S.H.M. equation in angular form `:`
`theta=theta_(m)sin (omega_(0)t+alpha)`
If at the initial moment` t=0, theta=0, ` then `alpha=0` . Then the above equation takes the form
`theta=theta_(m)sin omega_(0)t`
Let `v_(0)` be the velocity of the lower end of pendulum at `theta=0`, then form conserved of mechanical energy of oscillation
`E_(mean)=E_(extreme)` or `T_(mean )=U_(extreme)`
or, `(1)/(2) m v_(0)^(2)=mgl(1-cos theta_(m))`
Thus
`theta_(m)=cos^(-1)(1-(v_(0)^(2))/(2gl))=cos^(-1)[1-(0.22)^(2)/(2xx9.8xx0.8)]=4.5^(@)`
Thus the sought equation becomes
`theta= theta_(m) sin omega_(0)t=4.5^(@) sin 3.5 t`
`(c)` Let `theta_(0)` and `v_(0)` be the angular deviation and linear velocity at `t=0`.
As the mechanical energy of oscillation of the mathematical pendulum is conservation
`(1)/(2)m v_(0)^(2)+mgl (1-cos theta_(0))=mgl (1-cos theta_(m))`
or `(v_(0)^(2))/(2)=gl(cos theta_(0)-cos theta_(m))`
Thus `theta_(m)=cos^(-1){cos theta_(0)-(v_(0)^(2))/(2 gl)}=cos ^(-1){cos 3 ^(@)-((0.22)^(2))/(2xx9.8xx0.8)}=5.4^(@)`
Then from `theta=5.4^(@) sin (3.5 t+alpha)`, we see that ` sin alpha=(3)/(5.4)` and `cos alpha lt 0` because the velocity is directed towards the centre . Thus `alpha=(pi)/(2)+1.0` radians and we get the answer.