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For angular simple harmonic motion or physical pendulum torque is calculated about which point or line.( in reference to hc verma part 1 chapter 12 question 26)
in this kind of problems consider oscillation of centre of mass only that will make ur problem easier to solve let the pendulum be at some Φ angle with vertical axis, draw f.b.d ur self, balance the driving force with mg,i.e; F=mg sinΦ so a= F/m=g sinΦ if Φis very small then sinΦ ≈ Φ, a = gΦ =g(x/l) » a proportional to x then use time period formulae, t=2∏ √x/a .
in this kind of problems consider oscillation of centre of mass only that will make ur problem easier to solve
let the pendulum be at some Φ angle with vertical axis, draw f.b.d ur self,
balance the driving force with mg,i.e; F=mg sinΦ
so a= F/m=g sinΦ if Φis very small then sinΦ ≈ Φ,
a = gΦ =g(x/l) » a proportional to x
then use time period formulae, t=2∏ √x/a .
In a physical pendulum the pendulum is hinged about a point and it rotates about that point.........however in this case this point cant be CM ...so the torque has to be take about a point where it does not pass through the line of action......in case of a mathematical pendulum it moves about the pivoted point.................for deep concepts refer RESNICK AND HALLIDAY......pardon me if in any way im wrong.....
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