Question

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance x from the 50 cm mark. The period of oscillation is observed to be 2.9 s. Find the distance x.
m

Answer 1

answer) the period is given by

T=2piI/

I=moment of inertia=ML²/12 +Mx²

=Mgx

now using the above eqn we have

T²=2²pi²I/

2.9²=4pi²(ML²/12 +Mx²)/Mgx

2.9²gx=4pi²(1/12+x²)

82.418x=3.2865+39.4384x²

39.4384x²-82.418x+3.2865=0

its in the form of quadratic eqn

A=39.4384,B=-82.418,C=3.2865

x=-bb²-4ac/2a

x=-(-82.418)-82.418²-4*39.4348*3.2865/2*29.4384

x=2.05 m or x=0.041 m

we need value less than 1 m , so first value cancel out

so the answer is 0.041 m or 0.0407m